Question

Answer 1

Lời giải:
1. ĐKXĐ: $x\neq \pm 3$

\(A=\frac{x^2-x}{(x-3)(x+3)}-\frac{x+3}{(x-3)(x+3)}+\frac{x-3}{(x-3)(x+3)}\)

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

2. 

\(A=\frac{x+2}{x+3}=\frac{3}{4}\Leftrightarrow 4(x+2)=3(x+3)\)

$\Rightarrow x=1$ (tm) 

3.

\(A=\frac{x+2}{x+3}=1-\frac{1}{x+3}\)

Để $A$ nguyên thì $1-\frac{1}{x+3}$ nguyên 
$\Leftrightarrow \frac{1}{x+3}$ nguyên 

$\Rightarrow x+3$ là ước của $1$

$\Rightarrow x+3\in\left\{1; -1\right\}$

$\Rightarrow x\in\left\{-2; -4\right\}$ (tm)