3x2 solution. The impossible is possible, or how to solve the basic models of a Rubik's cube

Solving equations and inequalities with modulus often causes difficulties. However, if you understand well what it is the absolute value of a number, And how to correctly expand expressions containing a modulus sign, then the presence in the equation expression under the modulus sign, ceases to be an obstacle to its solution.

A little theory. Each number has two characteristics: absolute value number and its sign.

For example, the number +5, or simply 5, has a “+” sign and an absolute value of 5.

The number -5 has a "-" sign and an absolute value of 5.

The absolute values ​​of the numbers 5 and -5 are 5.

The absolute value of a number x is called the modulus of the number and is denoted by |x|.

As we see, the modulus of a number is equal to the number itself, if this number is greater than or equal to zero, and this number with opposite sign, if this number is negative.

The same applies to any expressions that appear under the modulus sign.

The module expansion rule looks like this:

|f(x)|= f(x) if f(x) ≥ 0, and

|f(x)|= - f(x), if f(x)< 0

For example |x-3|=x-3, if x-3≥0 and |x-3|=-(x-3)=3-x, if x-3<0.

To solve an equation containing an expression under the modulus sign, you must first expand a module according to the module expansion rule.

Then our equation or inequality becomes into two different equations existing on two different numerical intervals.

One equation exists on a numerical interval on which the expression under the modulus sign is non-negative.

And the second equation exists on the interval on which the expression under the modulus sign is negative.

Let's look at a simple example.

Let's solve the equation:

|x-3|=-x 2 +4x-3

1. Let's open the module.

|x-3|=x-3, if x-3≥0, i.e. if x≥3

|x-3|=-(x-3)=3-x if x-3<0, т.е. если х<3

2. We received two numerical intervals: x≥3 and x<3.

Let us consider into which equations the original equation is transformed on each interval:

A) For x≥3 |x-3|=x-3, and our wounding has the form:

Attention! This equation exists only on the interval x≥3!

Let's open the brackets and present similar terms:

and solve this equation.

This equation has roots:

x 1 =0, x 2 =3

Attention! since the equation x-3=-x 2 +4x-3 exists only on the interval x≥3, we are only interested in those roots that belong to this interval. This condition is satisfied only by x 2 =3.

B) At x<0 |x-3|=-(x-3) = 3-x, и наше уравнение приобретает вид:

Attention! This equation exists only on the interval x<3!

Let's open the brackets and present similar terms. We get the equation:

x 1 =2, x 2 =3

Attention! since the equation 3-x=-x 2 +4x-3 exists only on the interval x<3, нас интересуют только те корни, которые принадлежат этому промежутку. Этому условию удовлетворяет только х 1 =2.

So: from the first interval we take only the root x=3, from the second - the root x=2.

Solving exponential equations. Examples.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

What's happened exponential equation? This is an equation in which the unknowns (x's) and expressions with them are in indicators some degrees. And only there! It is important.

There you are examples of exponential equations:

3 x 2 x = 8 x+3

Note! In the bases of degrees (below) - only numbers. IN indicators degrees (above) - a wide variety of expressions with an X. If, suddenly, an X appears in the equation somewhere other than an indicator, for example:

this will already be an equation of mixed type. Such equations do not have clear rules for solving them. We will not consider them for now. Here we will deal with solving exponential equations in its purest form.

In fact, even pure exponential equations are not always solved clearly. But there are certain types of exponential equations that can and should be solved. These are the types we will consider.

Solving simple exponential equations.

First, let's solve something very basic. For example:

Even without any theories, by simple selection it is clear that x = 2. Nothing more, right!? No other value of X works. Now let's look at the solution to this tricky exponential equation:

What have we done? We, in fact, simply threw out the same bases (triples). Completely thrown out. And, the good news is, we hit the nail on the head!

Indeed, if in an exponential equation there are left and right the same numbers in any powers, these numbers can be removed and the exponents can be equalized. Mathematics allows. It remains to solve a much simpler equation. Great, right?)

However, let us remember firmly: You can remove bases only when the base numbers on the left and right are in splendid isolation! Without any neighbors and coefficients. Let's say in the equations:

2 x +2 x+1 = 2 3, or

twos cannot be removed!

Well, we have mastered the most important thing. How to move from evil exponential expressions to simpler equations.

"Those are the times!" - you say. “Who would give such a primitive lesson on tests and exams!?”

I have to agree. Nobody will. But now you know where to aim when solving tricky examples. It must be brought to the form where the same base number is on the left and right. Then everything will be easier. Actually, this is a classic of mathematics. We take the original example and transform it to the desired one us mind. According to the rules of mathematics, of course.

Let's look at examples that require some additional effort to reduce them to the simplest. Let's call them simple exponential equations.

Solving simple exponential equations. Examples.

When solving exponential equations, the main rules are actions with degrees. Without knowledge of these actions nothing will work.

To actions with degrees, one must add personal observation and ingenuity. Do we need the same base numbers? So we look for them in the example in explicit or encrypted form.

Let's see how this is done in practice?

Let us be given an example:

2 2x - 8 x+1 = 0

The first keen glance is at grounds. They... They are different! Two and eight. But it’s too early to become discouraged. It's time to remember that

Two and eight are relatives in degree.) It is quite possible to write:

8 x+1 = (2 3) x+1

If we recall the formula from operations with degrees:

(a n) m = a nm ,

this works out great:

8 x+1 = (2 3) x+1 = 2 3(x+1)

The original example began to look like this:

2 2x - 2 3(x+1) = 0

We transfer 2 3 (x+1) to the right (no one has canceled the elementary operations of mathematics!), we get:

2 2x = 2 3(x+1)

That's practically all. Removing the bases:

We solve this monster and get

This is the correct answer.

In this example, knowing the powers of two helped us out. We identified in eight there is an encrypted two. This technique (encoding common bases under different numbers) is a very popular technique in exponential equations! Yes, and in logarithms too. You must be able to recognize powers of other numbers in numbers. This is extremely important for solving exponential equations.

The fact is that raising any number to any power is not a problem. Multiply, even on paper, and that’s it. For example, anyone can raise 3 to the fifth power. 243 will work out if you know the multiplication table.) But in exponential equations, much more often it is not necessary to raise to a power, but vice versa... Find out what number to what degree is hidden behind the number 243, or, say, 343... No calculator will help you here.

You need to know the powers of some numbers by sight, right... Let's practice?

Determine what powers and what numbers the numbers are:

2; 8; 16; 27; 32; 64; 81; 100; 125; 128; 216; 243; 256; 343; 512; 625; 729, 1024.

Answers (in a mess, of course!):

5 4 ; 2 10 ; 7 3 ; 3 5 ; 2 7 ; 10 2 ; 2 6 ; 3 3 ; 2 3 ; 2 1 ; 3 6 ; 2 9 ; 2 8 ; 6 3 ; 5 3 ; 3 4 ; 2 5 ; 4 4 ; 4 2 ; 2 3 ; 9 3 ; 4 5 ; 8 2 ; 4 3 ; 8 3 .

If you look closely, you can see a strange fact. There are significantly more answers than tasks! Well, it happens... For example, 2 6, 4 3, 8 2 - that's all 64.

Let us assume that you have taken note of the information about familiarity with numbers.) Let me also remind you that to solve exponential equations we use all stock of mathematical knowledge. Including those from junior and middle classes. You didn’t go straight to high school, right?)

For example, when solving exponential equations, putting the common factor out of brackets often helps (hello to 7th grade!). Let's look at an example:

3 2x+4 -11 9 x = 210

And again, the first glance is at the foundations! The bases of the degrees are different... Three and nine. But we want them to be the same. Well, in this case the desire is completely fulfilled!) Because:

9 x = (3 2) x = 3 2x

Using the same rules for dealing with degrees:

3 2x+4 = 3 2x ·3 4

That’s great, you can write it down:

3 2x 3 4 - 11 3 2x = 210

We gave an example for the same reasons. So, what is next!? You can't throw out threes... Dead end?

Not at all. Remember the most universal and powerful decision rule everyone math tasks:

If you don’t know what you need, do what you can!

Look, everything will work out).

What's in this exponential equation Can do? Yes, on the left side it just begs to be taken out of brackets! The overall multiplier of 3 2x clearly hints at this. Let's try, and then we'll see:

3 2x (3 4 - 11) = 210

3 4 - 11 = 81 - 11 = 70

The example keeps getting better and better!

We remember that to eliminate grounds we need a pure degree, without any coefficients. The number 70 bothers us. So we divide both sides of the equation by 70, we get:

Oops! Everything got better!

This is the final answer.

It happens, however, that taxiing on the same basis is achieved, but their elimination is not possible. This happens in other types of exponential equations. Let's master this type.

Replacing a variable in solving exponential equations. Examples.

Let's solve the equation:

4 x - 3 2 x +2 = 0

First - as usual. Let's move on to one base. To a deuce.

4 x = (2 2) x = 2 2x

We get the equation:

2 2x - 3 2 x +2 = 0

And this is where we hang. The previous techniques will not work, no matter how you look at it. We'll have to pull out another powerful and universal method from our arsenal. It's called variable replacement.

The essence of the method is surprisingly simple. Instead of one complex icon (in our case - 2 x) we write another, simpler one (for example - t). Such a seemingly meaningless replacement leads to amazing results!) Everything just becomes clear and understandable!

So let

Then 2 2x = 2 x2 = (2 x) 2 = t 2

In our equation we replace all powers with x's by t:

Well, does it dawn on you?) Have you forgotten the quadratic equations yet? Solving through the discriminant, we get:

The main thing here is not to stop, as happens... This is not the answer yet, we need x, not t. Let's return to the X's, i.e. we make a reverse replacement. First for t 1:

That is,

One root was found. We are looking for the second one from t 2:

Hm... 2 x on the left, 1 on the right... Problem? Not at all! It is enough to remember (from operations with powers, yes...) that a unit is any number to the zero power. Any. Whatever is needed, we will install it. We need a two. Means:

That's it now. We got 2 roots:

This is the answer.

At solving exponential equations at the end sometimes you end up with some kind of awkward expression. Type:

Seven cannot be converted to two through a simple power. They are not relatives... How can we be? Someone may be confused... But the person who read on this site the topic “What is a logarithm?” , just smiles sparingly and writes down with a firm hand the absolutely correct answer:

There cannot be such an answer in tasks “B” on the Unified State Examination. There a specific number is required. But in tasks “C” it’s easy.

This lesson provides examples of solving the most common exponential equations. Let's highlight the main points.

Practical tips:

1. First of all, we look at grounds degrees. We are wondering if it is possible to make them identical. Let's try to do this by actively using actions with degrees. Don't forget that numbers without x's can also be converted to powers!

2. We try to bring the exponential equation to the form when on the left and on the right there are the same numbers in any powers. We use actions with degrees And factorization. What can be counted in numbers, we count.

3. If the second tip did not work, try using variable replacement. The result may be an equation that can be easily solved. Most often - square. Or fractional, which also reduces to square.

4. To successfully solve exponential equations, you need to know the powers of some numbers by sight.

As usual, at the end of the lesson you are invited to decide a little.) On your own. From simple to complex.

Solve exponential equations:

More difficult:

2 x+3 - 2 x+2 - 2 x = 48

9 x - 8 3 x = 9

2 x - 2 0.5x+1 - 8 = 0

Find the product of roots:

2 3's + 2 x = 9

Happened?

Well, then a very complex example (though it can be solved in the mind...):

7 0.13x + 13 0.7x+1 + 2 0.5x+1 = -3

What's more interesting? Then here's a bad example for you. Quite tempting for increased difficulty. Let me hint that in this example, what saves you is ingenuity and the most universal rule for solving all mathematical problems.)

2 5x-1 3 3x-1 5 2x-1 = 720 x

A simpler example, for relaxation):

9 2 x - 4 3 x = 0

And for dessert. Find the sum of the roots of the equation:

x 3 x - 9x + 7 3 x - 63 = 0

Yes Yes! This is a mixed type equation! Which we did not consider in this lesson. Why consider them, they need to be solved!) This lesson is quite enough to solve the equation. Well, you need ingenuity... And may seventh grade help you (this is a hint!).

Answers (in disarray, separated by semicolons):

1; 2; 3; 4; there are no solutions; 2; -2; -5; 4; 0.

Is everything successful? Great.

There is a problem? No problem! Special Section 555 solves all these exponential equations with detailed explanations. What, why, and why. And, of course, there is additional valuable information on working with all sorts of exponential equations. Not just these ones.)

One last fun question to consider. In this lesson we worked with exponential equations. Why didn’t I say a word about ODZ here? In equations, this is a very important thing, by the way...

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

The human intellect needs constant training no less than the body needs physical activity. The best way to develop and expand the abilities of this quality of the psyche is to solve crosswords and solve puzzles, the most famous of which, of course, is the Rubik's cube. However, not everyone manages to collect it. Knowledge of the diagrams and formulas for solving the assembly of this intricate toy will help you cope with this task.

What is a puzzle toy

A mechanical cube made of plastic, the outer edges of which consist of small cubes. The size of the toy is determined by the number of small elements:

• 2 x 2;
• 3 x 3 (the original version of the Rubik's cube was exactly 3 x 3);
• 4 x 4;
• 5 x 5;
• 6 x 6;
• 7 x 7;
• 8 x 8;
• 9 x 9;
• 10 x 10;
• 11 x 11;
• 13 x 13;
• 17 x 17.

Any of the small cubes can rotate in three directions along axes represented in the form of protrusions of a fragment of one of the three cylinders of the large cube. This way the structure can rotate freely, but small parts do not fall out, but hold on to each other.

Each face of the toy includes 9 elements, painted in one of six colors, located opposite each other in pairs. The classic combination of shades is:

• red opposite orange;
• white is opposite yellow;
• blue is opposite green.

However, modern versions can be painted in other combinations.

Today you can find Rubik's cubes of different colors and shapes.

This is interesting. The Rubik's cube even exists in a version for the blind. There, instead of color squares, there is a relief surface.

The goal of the puzzle is to arrange the small squares so that they form the edge of a large cube of the same color.

History of appearance

The idea of ​​the creation belongs to the Hungarian architect Erna Rubik, who, in fact, did not create a toy, but a visual aid for his students. The resourceful teacher planned to explain the theory of mathematical groups (algebraic structures) in such an interesting way. This happened in 1974, and a year later the invention was patented as a puzzle toy - future architects (and not only them) became so attached to the intricate and colorful manual.

The release of the first series of the puzzle was timed to coincide with the new year of 1978, but the toy came into the world thanks to entrepreneurs Tibor Lakzi and Tom Kremer.

This is interesting. Since its introduction, the Rubik's cube ("magic cube", "magic cube") has sold about 350 million copies worldwide, making the puzzle the number one most popular toy. Not to mention dozens of computer games based on this assembly principle.

The Rubik's Cube is an iconic toy for many generations

In the 80s, residents of the USSR became acquainted with the Rubik's cube, and in 1982, the first world championship in speed puzzle assembly - speedcubing - was organized in Hungary. Then the best result was 22.95 seconds (for comparison: a new world record was set in 2017: 4.69 seconds).

This is interesting. Fans of solving colorful puzzles are so attached to the toy that speed-assembling competitions alone are not enough for them. Therefore, in recent years, championships have appeared in solving puzzles with closed eyes, one hand, and feet.

What are the formulas for the Rubik's cube

To assemble a magic cube means to arrange all the small parts so that you get a whole face of the same color, you need to use God's algorithm. This term refers to a set of minimum actions that will solve a puzzle that has a finite number of moves and combinations.

This is interesting. In addition to the Rubik's cube, God's algorithm is applied to such puzzles as Meffert's pyramid, Taken, Tower of Hanoi, etc.

Since the magic Rubik's cube was created as a mathematical tool, its assembly is laid out according to formulas.

Solving a Rubik's cube is based on the use of special formulas

Important Definitions

In order to learn to understand the schemes for solving a puzzle, you need to become familiar with the names of its parts.

1. An angle is a combination of three colors. In the 3 x 3 cube there will be 3 of them, in the 4 x 4 version there will be 4, etc. The toy has 12 corners.
2. An edge represents two colors. There are 8 of them in a cube.
3. The center contains one color. There are 6 of them in total.
4. The faces, as already mentioned, are simultaneously rotating puzzle elements. They are also called “layers” or “slices”.

Values ​​in formulas

It should be noted that the assembly formulas are written in Latin - these are the diagrams that are widely presented in various manuals for working with the puzzle. But there are also Russified versions. The list below contains both options.

1. The front edge (front or façade) is the front edge, which is the color facing us [F] (or F - front).
2. The back face is the face that is centered away from us [B] (or B - back).
3. Right Face - the face that is on the right [P] (or R - right).
4. Left Face - the face that is on the left [L] (or L - left).
5. Bottom Face - the face that is at the bottom [H] (or D - down).
6. Top Face - the face that is at the top [B] (or U - up).

Photo gallery: parts of the Rubik's cube and their definitions

To explain the notation in the formulas, we use the Russian version - it will be clearer for beginners, but for those who want to move to the professional level of speedcubing, they cannot do without an international notation system in English.

This is interesting. The international notation system is adopted by the World Cube Association (WCA).

1. The central cubes are designated in the formulas by one lowercase letter - f, t, p, l, v, n.
2. Angular - three letters according to the name of the edges, for example, fpv, flni, etc.
3. Capital letters F, T, P, L, V, N indicate the elementary operations of rotating the corresponding face (layer, slice) of a cube 90° clockwise.
4. The designations F", T", P", L", V", N" correspond to the rotation of the faces by 90° counterclockwise.
5. The designations Ф 2, П 2, etc. indicate a double rotation of the corresponding face (Ф 2 = ФФ).
6. The letter C indicates the rotation of the middle layer. The subscript indicates which face should be viewed from in order to make this turn. For example, C P - from the right side, C N - from the bottom side, C "L - from the left side, counterclockwise, etc. It is clear that C N = C " B, C P = C " L and etc.
7. The letter O is a rotation (turn) of the entire cube around its axis. O F - from the side of the front edge clockwise, etc.

Recording the process (Ф "П") Н 2 (ПФ) means: rotate the front face counterclockwise by 90°, the same - the right edge, rotate the bottom edge twice (that is, 180°), rotate the right edge 90° along clockwise, rotate the front edge 90° clockwise.

Unknown

http://dedfoma.ru/kubikrubika/kak-sobrat-kubik-rubika-3x3x3.htm

It is important for beginners to learn to understand formulas

As a rule, the instructions for assembling a puzzle in classic colors recommend holding the puzzle with the yellow center facing up.

This advice is especially important for beginners.

This is interesting. There are sites that visualize formulas. Moreover, the speed of the assembly process can be set independently. For example, alg.cubing.net

How to solve a Rubik's puzzle

• There are two types of schemes:
• for newbies;

for professionals.

Their difference is in the complexity of the formulas, as well as the speed of assembly. For beginners, of course, instructions appropriate to their level of puzzle proficiency will be more useful. But after practice, they too will be able to fold the toy in 2–3 minutes.

How to solve a standard 3 x 3 cube

Let's start by solving the classic 3 x 3 Rubik's cube using a 7-step diagram.

The classic version of the puzzle is the 3 x 3 Rubik's Cube

This is interesting. The reverse process used to solve certain misplaced cubes is the reverse sequence of the action described by the formula. That is, the formula must be read from right to left, and the layers must be rotated counterclockwise if direct movement was specified, and vice versa: direct if the opposite is described.

1. We start by assembling the cross on the top edge. We lower the desired cube down by rotating the corresponding side face (P, T, L) and bring it to the front face using the operation H, N" or H 2. We finish the removal stage with a mirror rotation (reverse) of the same side face, restoring the original position of the affected rib cube of the upper layer. After this, we carry out operation a) or b) of the first stage. In case a) the cube has reached the front face so that the color of its front face coincides with the color of the front. In case b) the cube must not only be moved up, but also rotated. , so that it is correctly oriented, falling into place.

   

   Collecting the top line cross

2. The required corner cube is found (having the colors of the faces F, B, L) and, using the same technique described for the first stage, is brought to the left corner of the selected front face (or yellow). There are three possible orientations for this cube. We compare our case with the figure and apply one of the operations of the second stage a, beat c. The dots on the diagram mark the place where the desired cube should go. We find the remaining three corner cubes on the cube and repeat the described technique to move them to their places on the top face. Result: the top layer has been selected. The first two stages cause almost no difficulties for anyone: you can quite easily monitor your actions, since all attention is paid to one layer, and what is done in the remaining two is not at all important.

   

   Selecting the top layer

3. Our goal: to find the desired cube and first bring it down to the front face. If it is at the bottom, simply turn the bottom edge until it matches the color of the facade, and if it is in the middle layer, then you must first lower it down using any of operations a) or b), and then match it in color with the color of the facade edge and perform the third stage operation a) or b). Result: two layers are collected. The formulas given here are mirror ones in the full sense of the word. You can clearly see this if you place a mirror to the right or left of the cube (edge ​​facing you) and do any of the formulas in the mirror: we will see the second formula. That is, operations with the front, bottom, top (not involved here), and back (also not involved) faces change their sign to the opposite: it was clockwise, it became counterclockwise, and vice versa. And the left side changes from the right, and, accordingly, changes the direction of rotation to the opposite.

   

   We find the desired cube and bring it down to the front face

4. Operations that move the side cubes of one face without ultimately disturbing the order in the assembled layers lead to the goal. One of the processes that allows you to select all the side faces is shown in the figure. It also shows what happens to the other cubes of the face. By repeating the process, choosing another front face, you can put all four cubes in place. Result: The rib pieces are in place, but two of them, or even all four, may be oriented incorrectly. Important: before you start executing this formula, look at which cubes are already in place - they may be oriented incorrectly.

   

   If there is none or one, then we try to rotate the top face so that the two located on two adjacent side faces (fv+pv, pv+tv, tv+lv, lv+fv) fall into place, after which we orient the cube like this , as shown in the figure, and execute the formula given at this stage. If it is not possible to combine the parts belonging to adjacent faces by rotating the top face, then we perform the formula for any position of the cubes of the top face once and try again by rotating the top face to put in place 2 parts located on two adjacent side faces.

5. It is important to check the orientation of the cubes at this stage We take into account that the unfolded cube must be on the right side; in the figure it is marked with arrows (pv cube). Figures a, b, and c show possible cases of arrangement of incorrectly oriented cubes (marked with dots). Using the formula in case a), we perform an intermediate rotation B" to bring the second cube to the right side, and a final rotation B, which will return the top face to its original position, in case b) an intermediate rotation B 2 and the final one also B 2, and in case c) intermediate rotation B must be performed three times, after turning over each cube, and also completed with rotation B. Many people are confused by the fact that after the first part of the process (PS N) 4, the desired cube unfolds as it should, but the order in the assembled layers is disrupted. confusing and makes some people throw the almost completed cube halfway. Having performed an intermediate turn, not paying attention to the “breakage” of the lower layers, we perform operations (PS N) 4 with the second cube (the second part of the process), and everything falls into place.

   

   Result: the cross is assembled.

6. We put the corners of the last face in place using an 8-step process that is easy to remember - forward, rearranging the three corner pieces in a clockwise direction, and reverse, rearranging the three cubes in a counterclockwise direction. After the fifth stage, as a rule, at least one cube will sit in its place, albeit in the wrong direction. (If after the fifth stage none of the corner cubes are in their place, then we apply any of the two processes for any three cubes, after which exactly one cube will be in its place.). Result: All corner cubes are in place, but two (or maybe four) of them may be oriented incorrectly.

   

   Corner cubes sit in place

7. We repeat the sequence of turns PF"P"F many times. We rotate the cube so that the cube we want to expand is in the upper right corner of the facade. An 8-turn process (2 x 4 turns) will turn it 1/3 turn clockwise. If the cube has not yet oriented itself, we repeat the 8-move move again (in the formula this is reflected by the index “N”). We do not pay attention to the fact that the lower layers will become disordered. The figure shows four cases of incorrectly oriented cubes (they are marked with dots). In case a) an intermediate turn B and a final turn B are required, in case b) - an intermediate and final turn B 2, in case c) - turn B is performed after turning each cube to the correct orientation, and the final turn B 2, in case d) - intermediate rotation B is also performed after turning each cube to the correct orientation, and the final one in this case will also be rotation B. Result: the last face is assembled.

   

   Possible errors are shown by dots

Formulas for correcting the placement of cubes can be shown as follows.

Formulas for correcting incorrectly oriented cubes at the last stage

The essence of Jessica Friedrich's method

There are several ways to assemble the puzzle, but one of the most memorable is the one developed by Jessica Friedrich, a professor at the University of Binghamton (New York), who is developing techniques for hiding data in digital images. While still a teenager, Jessica became so interested in the cube that in 1982 she became the world champion in speedcubing and subsequently did not abandon her hobby, developing formulas for quickly assembling a “magic cube.” One of the most popular options for folding a cube is called CFOP - after the first letters of the four assembly steps.

Instructions:

1. We assemble a cross on the top face, which is made up of cubes on the edges of the bottom face. This stage is called Cross.
2. We assemble the bottom and middle layers, that is, the face on which the cross is located, and the intermediate layer, consisting of four side parts. The name of this step is F2L (First two layers).
3. We assemble the remaining edge, not paying attention to the fact that not all the parts are in place. The stage is called OLL (Orient the last layer), which translates as “orientation of the last layer.”
4. The last level - PLL (Permute the last layer) - consists of the correct placement of the cubes of the top layer.

Video instructions for the Friedrich method

The method that was proposed by Jessica Friedrich was so liked by speedcubers that the most advanced amateurs are developing their own methods to speed up the assembly of each of the stages proposed by the author.

Video: speeding up the assembly of the cross

Video: assembling the first two layers

Video: working with the last layer

Video: last level of assembly by Friedrich

2 x 2

A 2 x 2 Rubik's cube or mini Rubik's cube is also folded in layers, starting from the bottom level.

Mini cube is a light version of the classic puzzle

Beginner's instructions for easy assembly

1. We assemble the bottom layer so that the colors of the last four cubes match, and the remaining two colors are the same as the colors of the adjacent parts.
2. Let's start organizing the top layer. Please note that at this stage the goal is not to match the colors, but to put the cubes in their places. We start by determining the color of the top. Everything is simple here: this will be the color that did not appear in the bottom layer. Rotate any of the top cubes so that it gets to the position where the three colors of the element intersect. Having fixed the angle, we arrange the remaining elements. For this we use two formulas: one for changing diagonal cubes, the other for neighboring ones.
3. We complete the top layer. We carry out all operations in pairs: we rotate one corner and then the other, but in the opposite direction (for example, the first one clockwise, the second one counterclockwise). You can work with three angles at once, but in this case there will be only one combination: either clockwise or counterclockwise. Between rotations of the corners, rotate the top edge so that the corner being worked is in the upper right corner. If we are working with three corners, then place the correctly oriented one at the back left.

Formulas for rotating angles:

• (VFPV · P"V"F")² (5);
• V²F·V²F"·V"F·V"F"(6);
• VVF² · LFL² · VLV² (7).

To rotate three corners at once:

• (FVPV"P"F"V")² (8);
• FV·F"V·FV²·F"V² (9);
• V²L"V"L²F"L"F²V"F" (10).

Photo gallery: 2 x 2 cube assembly

Video: Friedrich method for 2 x 2 cube

Collecting the most difficult versions of the cube

These include toys with a number of parts from 4 x 4 and up to 17 x 17.

Cube models with many elements usually have rounded corners for ease of manipulation with the toy

This is interesting. A 19 x 19 version is currently being developed.

It should be remembered that they were created on the basis of a 3 x 3 cube, therefore the assembly is built in two directions.

1. We assemble the center so that the elements of the 3 x 3 cube remain.
2. We work according to the diagrams for assembling the initial version of the toy (most often cubers use Jessica Friedrich’s method).

4 x 4

This version is called "Rubik's Revenge".

Instructions:

The assembly of the 5 x 5, 6 x 6 and 7 x 7 models is similar to the previous one, only we take a larger number of cubes as the basis for the center.

Video: solving a Rubik's cube 5 x 5

Working on solving a 6 x 6 puzzle

This cube is quite inconvenient to work with: a large number of small parts require special attention. Therefore, we will divide the video instructions into four parts: for each stage of assembly.

Video: how to assemble the center of a 6 x 6 cube, part 1

Video: pairing edge elements in a 6 x 6 cube, part 2

Video: pairing four elements in a 6 x 6 puzzle, part 3

Video: final solving of the 6 x 6 Rubik's cube, part 4

Video: putting together a 7 x 7 puzzle

How to solve the pyramid puzzle

This puzzle is mistakenly considered a type of Rubik's cube. But in fact, Meffert’s toy, which is also called the “Japanese tetrahedron” or “Moldavian pyramid,” appeared several years earlier than the visual aid of the teacher-architect.

Meffert's pyramid is mistakenly called a Rubik's puzzle

To work with this puzzle, it is important to know its structure, because the operating mechanism plays a key role in assembly. The Japanese tetrahedron consists of:

• four axis elements;
• six ribs;
• four corners.

Each axle part has small triangles facing three adjacent faces. That is, each element can be rotated without the threat of it falling out of the structure.

This is interesting. There are 75,582,720 options for the arrangement of pyramid elements. Unlike a Rubik's cube, it's not that big of a deal. The classic version of the puzzle has 43,252,003,489,856,000 possible configurations.

Instructions and diagram

Video: a simple method for assembling a complete pyramid

Method for children

Using formulas and using methods to speed up the assembly will be too difficult for children just starting out with the puzzle. Therefore, the task of adults is to simplify the explanation as much as possible.

The Rubik's Cube is not only an opportunity to keep your child busy with a useful and interesting activity, but also a way to develop patience and perseverance.

This is interesting. It is better to start teaching children with the 3 x 3 model.

Instructions (3 x 3 cube):

1. We decide on the color of the top edge and take the toy so that the central cube of the desired color is at the top.
2. We assemble the top cross, but the second color of the middle layer was the same as the color of the side edges.
3. We set the corners of the top edge. Let's move on to the second layer.
4. We assemble the last layer, but start by restoring the sequence of the first ones. Then we set the corners so that they coincide with the central details of the edges.
5. We check the location of the middle parts of the last face, changing their location if necessary.

Solving a Rubik's cube in any of its variations is a great workout for the mind, a way to relieve stress and distract yourself. Even a child can learn to solve a puzzle using age-appropriate explanations. Gradually, you can master more intricate assembly methods, improve your own time indicators, and then you’re not far from speedcubing competitions. The main thing is persistence and patience.

Quadratic equations.

Quadratic equation- algebraic equation of general form

where x is a free variable,

a, b, c, are coefficients, and

Expression called a square trinomial.

Methods for solving quadratic equations.

1. METHOD : Factoring the left side of the equation.

Let's solve the equation x 2 + 10x - 24 = 0. Let's factorize the left side:

x 2 + 10x - 24 = x 2 + 12x - 2x - 24 = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as follows:

(x + 12)(x - 2) = 0

Since the product is zero, then at least one of its factors is zero. Therefore, the left side of the equation becomes zero at x = 2, and also when x = - 12. This means that the number 2 And - 12 are the roots of the equation x 2 + 10x - 24 = 0.

2. METHOD : Method for selecting a complete square.

Let's solve the equation x 2 + 6x - 7 = 0. Select a complete square on the left side.

To do this, we write the expression x 2 + 6x in the following form:

x 2 + 6x = x 2 + 2 x 3.

In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get a complete square, you need to add 3 2, since

x 2 + 2 x 3 + 3 2 = (x + 3) 2.

Let us now transform the left side of the equation

x 2 + 6x - 7 = 0,

adding to it and subtracting 3 2. We have:

x 2 + 6x - 7 = x 2 + 2 x 3 + 3 2 - 3 2 - 7 = (x + 3) 2 - 9 - 7 = (x + 3) 2 - 16.

Thus, this equation can be written as follows:

(x + 3) 2 - 16 =0, (x + 3) 2 = 16.

Hence, x + 3 - 4 = 0, x 1 = 1, or x + 3 = -4, x 2 = -7.

3. METHOD :Solving quadratic equations using the formula.

Let's multiply both sides of the equation

ax 2 + bx + c = 0, a ≠ 0

on 4a and sequentially we have:

4a 2 x 2 + 4abx + 4ac = 0,

((2ax) 2 + 2ax b + b 2) - b 2 + 4ac = 0,

(2ax + b) 2 = b 2 - 4ac,

2ax + b = ± √ b 2 - 4ac,

2ax = - b ± √ b 2 - 4ac,

Examples.

A) Let's solve the equation: 4x 2 + 7x + 3 = 0.

a = 4, b = 7, c = 3, D = b 2 - 4ac = 7 2 - 4 4 3 = 49 - 48 = 1,

D > 0, two different roots;

Thus, in the case of a positive discriminant, i.e. at

b 2 - 4ac >0, the equation ax 2 + bx + c = 0 has two different roots.

b) Let's solve the equation: 4x 2 - 4x + 1 = 0,

a = 4, b = - 4, c = 1, D = b 2 - 4ac = (-4) 2 - 4 4 1= 16 - 16 = 0,

D = 0, one root;

So, if the discriminant is zero, i.e. b 2 - 4ac = 0, then the equation

ax 2 + bx + c = 0 has a single root

V) Let's solve the equation: 2x 2 + 3x + 4 = 0,

a = 2, b = 3, c = 4, D = b 2 - 4ac = 3 2 - 4 2 4 = 9 - 32 = - 13, D< 0.

This equation has no roots.

So, if the discriminant is negative, i.e. b 2 - 4ac< 0 , the equation

ax 2 + bx + c = 0 has no roots.

Formula (1) of the roots of a quadratic equation ax 2 + bx + c = 0 allows you to find roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed verbally as follows: the roots of a quadratic equation are equal to a fraction whose numerator is equal to the second coefficient taken with the opposite sign, plus minus the square root of the square of this coefficient without quadruple the product of the first coefficient by the free term, and the denominator is double the first coefficient.

4. METHOD: Solving equations using Vieta's theorem.

As is known, the reduced quadratic equation has the form

x 2 + px + c = 0.(1)

Its roots satisfy Vieta’s theorem, which, when a =1 looks like

x 1 x 2 = q,

x 1 + x 2 = - p

From this we can draw the following conclusions (from the coefficients p and q we can predict the signs of the roots).

a) If the half-member q given equation (1) is positive ( q > 0), then the equation has two roots of equal sign and this depends on the second coefficient p. If R< 0 , then both roots are negative if R< 0 , then both roots are positive.

For example,

x 2 – 3x + 2 = 0; x 1 = 2 And x 2 = 1, because q = 2 > 0 And p = - 3< 0;

x 2 + 8x + 7 = 0; x 1 = - 7 And x 2 = - 1, because q = 7 > 0 And p= 8 > 0.

b) If a free member q given equation (1) is negative ( q< 0 ), then the equation has two roots of different sign, and the larger root will be positive if p< 0 , or negative if p > 0 .

For example,

x 2 + 4x – 5 = 0; x 1 = - 5 And x 2 = 1, because q= - 5< 0 And p = 4 > 0;

x 2 – 8x – 9 = 0; x 1 = 9 And x 2 = - 1, because q = - 9< 0 And p = - 8< 0.

Examples.

1) Let's solve the equation 345x 2 – 137x – 208 = 0.

Solution. Because a + b + c = 0 (345 – 137 – 208 = 0), That

x 1 = 1, x 2 = c/a = -208/345.

Answer: 1; -208/345.

2) Solve the equation 132x 2 – 247x + 115 = 0.

Solution. Because a + b + c = 0 (132 – 247 + 115 = 0), That

x 1 = 1, x 2 = c/a = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b = 2k is an even number, then the root formula

Example.

Let's solve the equation 3x2 - 14x + 16 = 0.

Solution. We have: a = 3, b = - 14, c = 16, k = - 7;

D = k 2 – ac = (- 7) 2 – 3 16 = 49 – 48 = 1, D > 0, two different roots;

Answer: 2; 8/3

IN. Reduced equation

x 2 + px + q= 0

coincides with a general equation in which a = 1, b = p And c = q. Therefore, for the reduced quadratic equation, the root formula is

Takes the form:

Formula (3) is especially convenient to use when R- even number.

Example. Let's solve the equation x 2 – 14x – 15 = 0.

Solution. We have: x 1.2 =7±

Answer: x 1 = 15; x 2 = -1.

5. METHOD: Solving equations graphically.

Example. Solve the equation x2 - 2x - 3 = 0.

Let's plot the function y = x2 - 2x - 3

1) We have: a = 1, b = -2, x0 = = 1, y0 = f(1) = 12 - 2 - 3 = -4. This means that the vertex of the parabola is the point (1; -4), and the axis of the parabola is the straight line x = 1.

2) Take two points on the x-axis that are symmetrical about the axis of the parabola, for example points x = -1 and x = 3.

We have f(-1) = f(3) = 0. Let’s construct points (-1; 0) and (3; 0) on the coordinate plane.

3) Through the points (-1; 0), (1; -4), (3; 0) we draw a parabola (Fig. 68).

The roots of the equation x2 - 2x - 3 = 0 are the abscissas of the points of intersection of the parabola with the x-axis; This means that the roots of the equation are: x1 = - 1, x2 - 3.