Question

(2х-4у-10)^2+(3х+у-1)^2=0

Answer 1

Ну вот как- то так... 

Answer 2

4x-8y-20+6x+2y-2=0
10x-6y-22=0
10x-6y=22   /2
5x-3y=11
5x=11-3y
x=11-3y/5

Answer 3

The equation you provided is a quadratic equation in terms of the variables x and y. Let's expand and simplify the equation:

(2x - 4y - 10)^2 + (3x + y - 1)^2 = 0

Expanding the squares, we have:

(4x^2 - 16xy + 40x + 16y^2 - 80y + 100) + (9x^2 + 6xy + x^2 + 6xy + y^2 - 2y + 9) = 0

Combining like terms, we get:

14x^2 - 4xy + 7y^2 - 72x - 82y + 109 = 0

This is a quadratic equation in x and y. However, it does not appear to have any real solutions since the sum of squares is equal to zero, which can only occur if each term within the parentheses is equal to zero individually.

Setting each term equal to zero:

14x^2 - 4xy + 7y^2 - 72x - 82y + 109 = 0 14x^2 - 4xy + 7y^2 = 72x + 82y - 109

Since the left side is always non-negative, and the right side is negative for some values, there are no real solutions for x and y that satisfy this equation.