Question

Упростить срочно пожалуйста ​

Answer 1

$x^2+x-12=a(x-x_1)(x-x_2)=(x-3)(x+4)\\x^2+x-12=0\\D=1+48=49=7^2\\\left.\begin{array}{cc}x^1=\frac{-1+7}2=3&x_2=\frac{-1-7}2=-4\end{array}\right.\\(\frac3{x+4}+\frac{6x}{x^2+x-12}-\frac1{x-3}):\frac{8x-13}{x^2-16}=\frac{3(x-3)+6x-(x+4)}{(x+4)(x-3)}\cdot\frac{(x+4)(x-4)}{8x-13}=\frac{3x-9+6x-x-4}{x-3}\cdot\frac{x-4}{8x-13}=\frac{8x-13}{x-3}\cdot\frac{x-4}{8x-13}=\frac{x-4}{x-3}$