Question

16x^2+y^2+|5x-y+3|=8xy

Answer 1

Выражение  представляется в виде:

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

Тк  квадрат и модуль неотрицательны,то такое возможно когда они оба равны нулю:

4x-y=0

5x-y+3=0

x+3=0

x=-3

y=-12

Ответ: x=-3  ; y=-12

Answer 2

The equation you provided is:

16x^2 + y^2 + |5x - y + 3| = 8xy

Let's break it down into separate cases based on the absolute value expression. First, let's consider the case when (5x - y + 3) is positive, which means we can remove the absolute value.

Case 1: (5x - y + 3) ≥ 0

In this case, the equation becomes:

16x^2 + y^2 + (5x - y + 3) = 8xy

Simplifying further:

16x^2 + y^2 + 5x - y + 3 = 8xy

Rearranging terms:

16x^2 + 5x - 8xy + y^2 - y + 3 = 0

Now, let's consider the case when (5x - y + 3) is negative, which means we need to negate the absolute value expression.

Case 2: (5x - y + 3) < 0

In this case, the equation becomes:

16x^2 + y^2 - (5x - y + 3) = 8xy

Simplifying further:

16x^2 + y^2 - 5x + y - 3 = 8xy

Rearranging terms:

16x^2 - 5x - 8xy + y^2 + y - 3 = 0

Therefore, the given equation has two cases, and depending on the values of (5x - y + 3), the equation will fall into one of these cases.