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