a. Rút gọn biểu thức A b. Tìm x để A>0 c. Tìm x thuộc Z để A nguyên dương

ĐKXĐ: $x\ne-3;x\ne2$

a) $A=\dfrac{x+2}{x+3}-\dfrac5{x^2+x-6}+\dfrac1{2-x}$

$=\dfrac{x+2}{x+3}-\dfrac5{(x-2)(x+3)}-\dfrac1{x-2}$

$=\dfrac{(x+2)(x-2)-5-(x+3)}{(x-2)(x+3)}$

$=\dfrac{x^2-4-5-x-3}{(x-2)(x+3)}$

$=\dfrac{x^2-x-12}{(x-2)(x+3)}=\dfrac{(x-4)(x+3)}{(x-2)(x-3)}=\dfrac{x-4}{x-2}$

b) $A>0$

$\Rightarrow\left[\begin{array}{I}\begin{cases}x-4>0\\x-2>0\end{cases}\\\begin{cases}x-4<0\\x-2<0\end{cases}\end{array}\right.\Leftrightarrow\left[\begin{array}{I}x>4\\x<2(x\ne-3)\end{array}\right.$

c) $A>0\Leftrightarrow\left[\begin{array}{I}x>4\\x<2(x\ne-3)\end{array}\right.$

$A\in\mathbb Z\Rightarrow \dfrac{x+4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac2{x-2}\in\mathbb Z$

$\Rightarrow 2$ $\vdots$ $(x-2)$

$\Rightarrow x-2\in\{\pm1;\pm2\}$

$\Rightarrow x\in\{3;1;4;0\}$ so sánh với điều kiện

$\Rightarrow x\in\{1;0\}$.