Question

(x+3y-2)²+x²-10xy+25y²=0

Answer 1

(x+3y-2)²+x²-10x+25=0
(x+3y-2)²+(x-5)²=0
Так как (x+3y-2)²≥0   и  (x-5)²≥0    ⇒
Получаем систему уравнений:
(x+3y-2)²=0      x+3y-2=0      5+3y-2=0       3y+3=0   3y=-3  |÷3     y=-1..
(x-5)²=0            x-5=0             x=5.
Ответ: x=5    y=-1.

Answer 2

To solve the quadratic equation given, we'll first expand the equation and then simplify it. Let's work step-by-step:

The equation is: (x + 3y - 2)² + x² - 10xy + 25y² = 0

Step 1: Expand the square term (x + 3y - 2)²

(x + 3y - 2)² = (x + 3y - 2)(x + 3y - 2) = x(x + 3y - 2) + 3y(x + 3y - 2) - 2(x + 3y - 2) = x(x) + x(3y) + x(-2) + 3y(x) + 3y(3y) + 3y(-2) - 2(x) - 2(3y) - 2(-2) = x² + 3xy - 2x + 3xy + 9y² - 6y - 2x - 6y + 4 = x² + 6xy + 9y² - 4x - 12y + 4

Step 2: Substitute the expanded square term into the original equation

(x + 3y - 2)² + x² - 10xy + 25y² = 0 (x² + 6xy + 9y² - 4x - 12y + 4) + x² - 10xy + 25y² = 0

Step 3: Combine like terms

2x² - 4x + 10xy + 34y² - 12y + 4 = 0

Now, we have a quadratic equation in terms of x and y:

2x² + 10xy + 34y² - 4x - 12y + 4 = 0

This equation is not factorable into simple linear terms, so we may have to use numerical methods or complete the square to solve for specific values of x and y. If you have any specific values or constraints for x and y, please let me know, and I can assist you further with the solution.