Question

1){x3+y3=1 {x+y=1 2){x2+xy=12 {xy-y2=2 3)x3-y3=8 {x-y=2 Смотреть ответ Kislinki Kislinki 3х-3у=1 х=1-у 3(1-у)-3у=1 х=1-у 3-3у-3у+1=о 4-6у=о -6у=-4. :-6 у= 2:3(дробь) так типо ? х=1:3(дробь)

Answer 1

It seems like you have a system of equations that you are trying to solve. Let's go through each set of equations one by one:

1. x^3 + y^3 = 1 x + y = 1

2. x^2 + xy = 12 xy - y^2 = 2

3. x^3 - y^3 = 8 x - y = 2

For each set of equations, we'll solve them step by step:

1. x^3 + y^3 = 1 x + y = 1

We can solve the second equation for x: x = 1 - y Now substitute this value of x into the first equation:

(1 - y)^3 + y^3 = 1

Expand (1 - y)^3 using the binomial formula:

1^3 - 3 * 1^2 * y + 3 * 1 * y^2 - y^3 + y^3 = 1

Simplify:

1 - 3y + 3y^2 - y^3 + y^3 = 1

The y^3 terms cancel out:

1 - 3y + 3y^2 = 1

Now, let's move all terms to one side of the equation:

3y^2 - 3y = 0

Factor out common terms:

3y(y - 1) = 0

Now, set each factor to zero and solve for y:

1. y = 0
2. y - 1 = 0 => y = 1

Now, we have two potential values for y. Let's find the corresponding values for x:

For y = 0: x = 1 - y = 1 - 0 = 1

For y = 1: x = 1 - y = 1 - 1 = 0

So, for the first set of equations, we have two solutions: (x = 1, y = 0) and (x = 0, y = 1).

Let's move on to the second set of equations:

2. x^2 + xy = 12 xy - y^2 = 2

We can rearrange the first equation to solve for x:

x^2 + xy - 12 = 0

Now, let's use the quadratic formula to find the values of x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = y, and c = -12.

Now, the second equation is:

xy - y^2 = 2

We can rearrange this to solve for x:

x = (y^2 + 2) / y

Now, we can equate the two expressions for x:

(-y ± √(y^2 - 4(-12))) / 2 = (y^2 + 2) / y

Simplify:

(-y ± √(y^2 + 48)) / 2 = (y^2 + 2) / y

Now, let's solve this equation for y:

(-y ± √(y^2 + 48)) / 2 = (y^2 + 2) / y

y^2 * (-y ± √(y^2 + 48)) = 2 * (y^2 + 2)

y^2 * (-y ± √(y^2 + 48)) = 2y^2 + 4

y^2 * (-1 ± √(1 + 48)) = 2y^2 + 4

y^2 * (-1 ± √49) = 2y^2 + 4

Now, we have two cases:

Case 1: (-1 + √49) = 6

y^2 * 6 = 2y^2 + 4

6y^2 = 2y^2 + 4

4y^2 = 4

y^2 = 1

y = ±√1

y = ±1

Now, find the corresponding values of x using the expression we found earlier:

x = (y^2 + 2) / y

For y = 1: x = (1^2 + 2) / 1 x = 3

For y = -1: x = ((-1)^2 + 2) / -1 x = -1

Case 2: (-1 - √49) = -8

y^2 * (-8) = 2y^2 + 4

-8y^2 = 2y^2 + 4

-10y^2 = 4

y^2 = -4/10

y = ±√(-4/10)

Since the square root of a negative number is not real, there are no real solutions for this case.

So, for the second set of equations, we have two solutions: (x = 3, y = 1) and (x = -1, y = 1).

Finally, let's solve the third set of equations:

3. x^3 - y^3 = 8 x - y = 2

We can solve the second equation for x: x = 2 + y Now substitute this value of x into the first equation:

(2 + y)^3 - y^3 = 8

Expand (2 + y)^3 using the binomial formula:

8 + 12y + 6y^2 + y^3 - y^3 = 8

Simplify:

6y^2 + 12y = 0

Factor out common terms:

6y(y + 2) = 0

Now, set each factor to zero and solve for y:

1. y = 0
2. y + 2 = 0 => y = -2

Now, find the corresponding values of x using the expression we found earlier:

For y = 0: x = 2 + y = 2 + 0 = 2

For y = -2: x = 2 + y = 2 + (-2) = 0

So, for the third set of equations, we have two solutions: (x = 2, y = 0) and (x = 0, y = -2).

In summary, the solutions to the three sets of equations are as follows:

1. (x = 1, y = 0) and (x = 0, y = 1)
2. (x = 3, y = 1) and (x = -1, y = 1)
3. (x = 2, y = 0) and (x = 0, y = -2)

Please note that some of these solutions involve fractions or complex numbers, so make sure to double-check the calculations if needed.