Cho biểu thức $P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}$ . Tìm x để $P=\frac{-3}{4}$ .

Ta có $P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}$

$=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}$

ĐKXĐ: $\left\{\begin{array}{l}x+3\ne 0\\ 2-x\ne 0\end{array}\right.\Rightarrow \left\{\begin{array}{l}x\ne -3\\ x\ne 2\end{array}\right.$

$P=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}$

$=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}$

$=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}$

$=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}$