Question

Решите систему уравнений: x^2+y^2=2 xy=1

Answer 1

x²+y²=2  и xy=1;

x²+ 2xy + y² = 2  + 2xy и xy=1;

(x + y)² = 2  + 2xy и xy=1;

(x + y)² = 2  + 2 и xy=1;

(x + y)² = 4 и xy=1;

1) x + y = 2 и  xy=1;          или            2) x + y = -2 и ху = 1;

x₁ = 1 и  y₁=1;                                            x₂ = -1 и у₂ = -1;

(1; 1)                                                             (-1; -1)

Ответ: (1; 1), (-1;-1).

Answer 2

{х²+у²=2
{ху=1

х²+у²+2ху-2ху=2
(х+у)²-2=2
(х+у)²=2²

х+у=±2

1){х+у=2
{ху=1
х=2-у
(2-у)*у=1
-у²+2у-1=0
у²-2у+1=0
(у-1)²=0
у-1=0;у=1;х=2-у=2-1=1

ответ (1;1)

2){х+у=-2
{ху=1
х=-2-у
(-2-у)*у=1
-2у-у²-1=0
у²+2у+1=0
(у+1)²=0
у+1=0
у=-1
х=-2-у=-2+1=-1
ответ (-1;-1)

итого (1;1);(-1;-1)

Answer 3

To solve the system of equations:

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

We can use substitution to find the values of x and y.

Step 1: Solve equation 2 for x: xy = 1 x = 1/y

Step 2: Substitute the value of x from equation 2 into equation 1: (1/y)^2 + y^2 = 2 1/y^2 + y^2 = 2

Step 3: Multiply the entire equation by y^2 to get rid of the fraction: 1 + y^4 = 2y^2

Step 4: Rearrange the equation to form a quadratic equation: y^4 - 2y^2 + 1 = 0

Step 5: Factor the quadratic equation: (y^2 - 1)^2 = 0

Step 6: Solve for y: y^2 - 1 = 0 y^2 = 1

y = ±√1 y = ±1

Step 7: Find the corresponding values of x using the expression we found for x in Step 1: If y = 1, then x = 1/1 = 1 If y = -1, then x = 1/-1 = -1

So, the solutions to the system of equations are: x = 1, y = 1 x = -1, y = -1