Решите номер 32.Есть вложение.

Решите задачу:

$\frac{3x^2-2x}{6-7x-3x^2} = \frac{x(3x-2)}{-3(x- \frac{2}{3})(x+3)} } =\frac{x(3x-2)}{-(3x-2)(x+3)} } = -\frac{x}{x+3 } \\\ 6-7x-3x^2=0 \\\ 3x^2+7x-6=0 \\\ D=49+4\cdot3\cdot6=121 \\\ x_1= \frac{-7-11}{6} =-3 \\\ x_2= \frac{-7+11}{6} = \frac{2}{3}$

$\frac{7x^2-x}{2-13x-7x^2} = \frac{7x^2-x}{-7(x- \frac{1}{7})(x+2)} =\frac{x(7x-1)}{-(7x- 1)(x+2)} = -\frac{x}{x+2} \\\ 2-13x-7x^2=0 \\ 7x^2+13x-2=0 \\\ D=169+2\cdot7\cdot4=225 \\\ x_1= \frac{-13-15}{14} =-2 \\\ x_2= \frac{-13+15}{14} = \frac{1}{7}$