Giúp e bài 1 cần gấpppppppppppppp

Giải thích các bước giải:

a.Ta có:

$C=(\dfrac{2+x}{2-x}+\dfrac{4x^2}{4-x^2}-\dfrac{2-x}{2+x}):\dfrac{x^2-x}{2x-x^2}$

$\to C=(-\dfrac{2+x}{x-2}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}):\dfrac{x^2-x}{2x-x^2}$

$\to C=(-\dfrac{\left(2+x\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}):\dfrac{x(x-1)}{x(2-x)}$

$\to C=(\dfrac{-\left(2+x\right)^2-4x^2-\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}):\dfrac{x-1}{2-x}$

$\to C=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}:\dfrac{x-1}{-(x-2)}$

$\to C=\dfrac{-4x(x+2)}{\left(x-2\right)\left(x+2\right)}\cdot \dfrac{-(x-2)}{x-1}$

$\to C=-\dfrac{4x}{x-2}\cdot \dfrac{-(x-2)}{x-1}$

$\to C=\dfrac{4x}{x-1}$

b.Để $C<4$

$\to \dfrac{4x}{x-1}<4$

$\to \dfrac{4x}{x-1}-4<0$

$\to \dfrac{4x-4(x-1)}{x-1}<0$

$\to \dfrac{4}{x-1}<0$

$\to x-1<0$

$\to x<1$

c.Để $C\in Z$

$\to \dfrac{4x}{x-1}\in Z $

$\to \dfrac{4(x-1)+4}{x-1}\in Z $

$\to \dfrac4{x-1}\in Z$

$\to 4\quad\vdots\quad x-1$

$\to x-1\in U(4)$

$\to x-1\in\{1,2,4,-1, -2, -4\}$

$\to x\in\{2, 3, 5, 0, -1, -3\}$

 Mà $x\ne \pm2, x\ne 0$

$\to x\in\{ 3, 5,  -1, -3\}$