Làm giúp mình với mọi người ơi

Giải thích các bước giải:

Bài 10:

a.ĐKXĐ: $x\ne \pm2, x\ne3$

Ta có:

$A=(\dfrac1{x+2}+\dfrac5{x-2}+\dfrac4{x^2-4}):\dfrac6{x+3}$ 

$\to A=(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}+\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{4}{\left(x+2\right)\left(x-2\right)})\cdot\dfrac{x+3}6$

$\to A=\dfrac{x-2+5\left(x+2\right)+4}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+3}6$

$\to A=\dfrac{6x+12}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+3}6$

$\to A=\dfrac{6}{x-2}\cdot\dfrac{x+3}6$

$\to A=\dfrac{x+3}{x-2}$

b.Để $A\in Z$

$\to x+3\quad\vdots\quad x-2$

$\to x-2+5\quad\vdots\quad x-2$

$\to 5\quad\vdots\quad x-2$

$\to x-2\in U(5)$

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

$\to x\in\{3, 7, 1, -3\}$

Do $x\ne -3$

$\to x\in\{3, 7, 1\}$

Bài 11:

a.ĐKXĐ: $x\ne \pm2$

Ta có:

$A=(\dfrac{x+2}{x-2}-\dfrac1{x+2}-\dfrac{x-4}{4-x^2}):\dfrac1{x^2-4}$

$\to A=(\dfrac{x+2}{x-2}-\dfrac1{x+2}+\dfrac{x-4}{x^2-4})\cdot (x^2-4)$

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

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

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

$\to A=x^2+4x+2$

b.Để $A=14$

$\to x^2+4x+2=14$

$\to x^2+4x+4=16$

$\to (x+2)^2=16$

$\to x+2=\pm4$

$\to x\in\{2, -6\}$

Do $x\ne\pm2$

$\to x=-6$