Question

Решить системы: 1) x+y=1 x^3+y^3=2 2) x+y=1 x^4+y^4=2

Answer 1

Первая система. Решение задания приложено

Answer 2

To solve each system of equations, we'll use the method of substitution. Let's solve them one by one:

1. x + y = 1 x^3 + y^3 = 2

To solve this system, we can express y in terms of x from the first equation and then substitute it into the second equation.

From the first equation, we have: y = 1 - x

Now, substitute y in the second equation: x^3 + (1 - x)^3 = 2

Expand the cubic expression: x^3 + (1 - 3x + 3x^2 - x^3) = 2

Combine like terms: 3x^2 - 3x + 1 = 2

Now, bring all terms to one side of the equation: 3x^2 - 3x - 1 = 0

Now we have a quadratic equation in terms of x. We can solve it using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -3, and c = -1.

x = (3 ± √(9 + 12)) / 6 x = (3 ± √21) / 6

So the two possible values for x are: x₁ = (3 + √21) / 6 x₂ = (3 - √21) / 6

Now, to find the corresponding values of y, we'll use y = 1 - x:

For x₁: y₁ = 1 - ((3 + √21) / 6)

For x₂: y₂ = 1 - ((3 - √21) / 6)

Now we have two pairs of solutions: (x₁, y₁) and (x₂, y₂).

2. x + y = 1 x^4 + y^4 = 2

Similarly, we'll express y in terms of x from the first equation and then substitute it into the second equation.

From the first equation, we have: y = 1 - x

Now, substitute y in the second equation: x^4 + (1 - x)^4 = 2

Expand the quartic expression: x^4 + (1 - 4x + 6x^2 - 4x^3 + x^4) = 2

Combine like terms: 6x^2 - 4x + 1 = 2

Now, bring all terms to one side of the equation: 6x^2 - 4x - 1 = 0

Now we have a quadratic equation in terms of x. We can solve it using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 6, b = -4, and c = -1.

x = (4 ± √(16 + 24)) / 12 x = (4 ± √40) / 12 x = (4 ± 2√10) / 12 x = (2 ± √10) / 6

So the two possible values for x are: x₁ = (2 + √10) / 6 x₂ = (2 - √10) / 6

Now, to find the corresponding values of y, we'll use y = 1 - x:

For x₁: y₁ = 1 - ((2 + √10) / 6)

For x₂: y₂ = 1 - ((2 - √10) / 6)

Now we have two pairs of solutions: (x₁, y₁) and (x₂, y₂).