Question

(5x+y-6)^2+(3x-y-2)^4=0

Answer 1

Рассмотрим это уравнение подробнее. Т.к. каждая скобка стоит в чётной степени, то она не может быть отрицательная. Чтобы два неотрицательных выражения давали в сумме 0, нужно чтобы каждое из них обращалось в ноль. Запишем систему:

 $\left \{ {{5x+y-6=0;} \atop {3x-y-2=0}} \right. \\ \\ 8x=8; \\ x=1.\\ y=6-5x=6-5=1.\\\\ Answer: x=1, y=1.$ 

Answer 2

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The equation you entered is a biquadratic equation, which means it has four possible solutions for x and y. To solve it, you need to expand the brackets, collect the terms, and factorize the polynomial. Then, you can use the zero product property to find the values of x and y that satisfy the equation.

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