Cho biểu thức 

- Hướng dẫn giải

a) Điều kiện xác định:$\left\{\begin{array}{l}a+3\ne 0\\ a-2\ne 0\\ a\ne 0\end{array}\right.$ hay $a\ne \left\{-3;0;2\right\}$

b) 

$\begin{array}{l}P=\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a}{a^2-2a}\\ =\frac{a+2}{a+3}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{1}{a-2}\\ =\frac{\left(a+2\right)\left(a-2\right)}{\left(a+3\right)\left(a-2\right)}-\frac{5}{\left(a+3\right)\left(a-2\right)}-\frac{a+3}{\left(a+3\right)\left(a-2\right)}\\ =\frac{\left(a+2\right)\left(a-2\right)-5-\left(a+3\right)}{\left(a+3\right)\left(a-2\right)}\\ =\frac{a^2-4-5-a-3}{\left(a+3\right)\left(a-2\right)}\\ =\frac{a^2-a-12}{\left(a+3\right)\left(a-2\right)}\\ =\frac{\left(a+3\right)\left(a-4\right)}{\left(a+3\right)\left(a-2\right)}\\ =\frac{a-4}{a-2}\end{array}$

Vậy $P=\frac{a-4}{a-2}$