Question

Помогите пажалуйста,39.19

Answer 1

1)
$\frac{y-25}{5y - 25} + \frac{3y + 5}{y^2-5y} = \frac{y - 25}{5(y-5)} + \frac{3y+5}{y(y-5)} = \frac{y(y-25) + 5(3y+5)}{5y(y-5)} = \\ \\ = \frac{y^2-25y +15y +25}{5y(y-5)} = \frac{y^2-10y +5^2}{5y(y-5)} = \frac{(y-5)^2}{5y(y-5)} = \frac{y-5}{5y} = \\ \\ = \frac{y}{5y} - \frac{5}{5y} = \frac{1}{5} - \frac{1}{y} = 0.2 - \frac{1}{y} \\ \\ y= 2.5 \\ 0.2 - \frac{1}{2.5} = 0.2 - 0.4 = - 0.2$

2)
$\frac{2}{y^2-yx} - \frac{2}{yx -x^2} = \frac{2}{y(y-x)} - \frac{2}{x(y-x)} = \frac{2x-2y}{xy(y-x)} = \frac{-2(y-x)}{xy(y-x)} = \\ \\ = \frac{-2}{xy} = - \frac{2}{xy} \\ \\ x=2 ; y = -3 \\ - \frac{2}{2*(-3)} = - (- \frac{1}{3} ) = \frac{1}{3}$