Выполните действие: 4+x² 4x ------- + ------- = (2-x)³ (x-2)³ 3(x-7)²+8x-56= -x²+b²+2+1=

Решите задачу:

$1)\frac{4+x^2}{\left(2-x\right)^3}+\frac{4x}{\left(x-2\right)^3}=\frac{x^2+4}{-\left(x-2\right)^3}+\frac{4x}{\left(x-2\right)^3}=-\frac{x^2+4}{\left(x-2\right)^3}+\frac{4x}{\left(x-2\right)^3}= \\ \\ =\frac{-\left(x^2+4\right)+4x}{\left(-2+x\right)^3}=-\frac{\left(x-2\right)^2}{\left(x-2\right)^3}=-\frac{1}{x-2} \\ \\ \\ 2)3\left(x-7\right)^2+8x-56=3\left(x^2-14x+49\right)+8x-56= \\ \\= 3x^2-42x+147+8x-56=3x^2-34x+91 \\ \\ \\ 3)-x^2+b^2+2+1=b^2+3-x^2$