Последний пожалуйста,срочноо

Решите задачу:

$1- \frac{2x^2+9x-5}{x^2+x-6} : \frac{2x^2-3x+1}{x^2+5x+6} =1- \frac{2x^2+10x-x-5}{x^2+3x-2x-6} * \frac{x^2+5x+6}{2x^2-3x+1} =$
$1- \frac{2x(x+5)-(x+5)}{x(x+3)-2(x+3)} * \frac{x^2+3x+2x+6}{2x^2-x-2x+1} =1- \frac{(2x-1)(x+5)}{(x-2)(x+3)} * \frac{x(x+3)+2(x+3)}{x(2x-1)-(2x-1)} =$
$1- \frac{(2x-1)(x+5)}{(x-2)(x+3)} * \frac{(x+2)(x+3)}{(x-1)(2x-1)} =1- \frac{x+5}{x-2} * \frac{x+2}{x-1} =1- \frac{(x+5)(x+2)}{(x-2)(x-1)} =$
$1- \frac{x^2+2x+5x+10}{(x-2)(x-1)} =1- \frac{x^2+7x+10}{(x-20(x-1)} = \frac{(x-2)(x-1)-(x^2+7x+10)}{(x-2)(x-1)} =$
$\frac{x^2-x-2x+2-x^2-7x-10}{(x-2)(x-1)} = \frac{-10x-8}{(x-20(x-1)} .$