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) x² + 3x / x - 4 = x² / 4 - x
(x² + 3x) • (4 - x) = x² • (x - 4)
(x² - x³ + 12x - 3x² - x³ + 4x² = 0
5x² - 2x³ + 12x = 0
x • (5x - 2x² + 12) = 0
x • (-2x² + 5x + 12) = 0
x • (-2x² + 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/y²+3
5/y+3 - 3/y - 2-y/y²+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