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 and you are trying to solve it. Let's go through each of the systems and find the values of x and y.

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

This is a system of nonlinear equations, and solving it directly might be a bit challenging. Let's try a different approach to solve it.

Let's first solve the second equation for y in terms of x: y = 1 - x

Now, substitute this value of y into the first equation: x^3 + (1 - x)^3 = 1

Expand the equation: x^3 + (1 - x)(1 - x)(1 - x) = 1 x^3 + (1 - 3x + 3x^2 - x^3) = 1 x^3 - 3x^2 + 3x - 1 = 0

Now, we have a cubic equation. Solving cubic equations can be a bit more complicated, but we can observe that x = 1 is one of its solutions (by substituting x = 1 into the equation above).

Now, we can use synthetic division or other methods to find the quadratic factor of the equation:

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

Now, we have two more possible solutions from the quadratic factor: x^2 - 2x + 1 = 0

Using the quadratic formula, we find: x = (2 ± √2^2 - 411) / 2 x = (2 ± √0) / 2 x = 1

So, in this case, x = 1 is the only real solution. Now, we can find y using the second equation: y = 1 - x y = 1 - 1 y = 0

Therefore, the solution to the first system is x = 1 and y = 0.

Let's move on to the second system:

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

We can use substitution to solve this system as well. From the second equation, we can express y in terms of x: xy - y^2 = 2 y(x - y) = 2 y = 2 / (x - y)

Now, substitute this value of y into the first equation: x^2 + x(2 / (x - y)) = 12

Now, simplify the equation: x^2 + 2x / (x - y) = 12

Since we don't have a specific value for x, we can't find the exact solution yet. However, if you provide a specific value for x or if there is more information missing, we might be able to proceed with the solution.

Finally, let's take a look at the third system:

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

This system can be solved easily by recognizing the difference of cubes in the first equation:

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

Since we also have the second equation, x - y = 2, we can substitute it into the above equation: 2(x^2 + xy + y^2) = 8 x^2 + xy + y^2 = 4

We can further simplify this equation using the fact that x - y = 2: x^2 + xy + y^2 = (x - y)^2 + xy = 4

Now, we have x - y = 2 and x^2 + xy + y^2 = 4. By substituting x - y = 2 into the equation, we get:

2^2 + xy = 4 4 + xy = 4 xy = 0

Now we have the value of xy, but we still need to find the values of x and y. It seems that more information might be missing from this system as well.

In conclusion, we have partially solved the first system with x = 1 and y = 0. However, the other systems need more information to be solved completely. If there are additional constraints or missing equations, please provide them so we can continue with the solutions.