Question

Помогите, пожалуйста, с математикой. Очень срочно.

Answer 1

$\displaystyle\tt 1. \ \ \ \frac{(x+y)^2}{x^2-y^2}\cdot\frac{3x}{6x+6y} =\frac{(x+y)^2}{(x-y)(x+y)}\cdot\frac{3x}{6(x+y)}=\frac{x}{2(x-y)}\\\\\\ 2. \ \ \ \frac{x}{2(x-y)}=\frac{-4,5}{2(-4,5-0,5)}=\frac{-4,5}{2\cdot(-5)}=\frac{-4,5}{-10}=0,45\\\\\\ 3. \\\\ \frac{x}{2(x-3)}=0,4; \ \ \ x\neq 3\\\\ x=0,4\cdot2(x-3)\\\\x=0,8x-2,4\\\\x -0,8x=-2,4\\\\0,2x=-2,4\\\\x=-12$