1.Решите уравнение:а) (x^2+3x)/(x-4)=x^2/(4-x)б) 5/(y+3)-3/y=(2-y)/(y^2+3y)

А) x + 3x / x - 4 = x / 4 - x
(x + 3x) (4 - x) = x (x - 4)
(x - x + 12x - 3x - x + 4x = 0
5x - 2x + 12x = 0
x (5x - 2x + 12) = 0
x (-2x + 5x + 12) = 0
x (-2x + 8x - 3x + 12) = 0
x (-2x (x - 4) - 3 (x - 4)) = 0
x (-2x - 3) (x - 4) = 0
x2 = 0
x1 = -3/2

b) 5/y+3 - 3/y = 2-y/y+3
5/y+3 - 3/y - 2-y/y+3y = 0
5/y+3 - 3/y - 2-y/y(y+3) = 0
5y-3(y+3)-(2-y)/y(y+3) = 0
5y-3y-9-2+y/y(y+3) = 0
3y-11/y(y+3) = 0
3y - 11 = 0
3y = 11
y = 11/3
y = 3 2/3