Question

Сократите дробь: подробно
(6-3x)^2/3x^2+3x-18
Помогите

Answer 1

$\frac{ (6-3x)^{2} }{3 x^{2} +3x-18}= \frac{(3( 2-x))^{2} }{3( x^{2} +x-6)}= \frac{ 3^{2} ( 2-x)^{2} }{3( x^{2} +3x-2x-6)}= \frac{3(- (x-2))^{2} }{x(x+3)-2(x+3)}= \frac{3( x-2)^{2} }{(x+3)(x-2)} = \\ \\ \\ = \frac{3(x-2)}{x+3}= \frac{3x-6}{x+3}$