a: ĐKXĐ: x∉{3;-3;-2}
\(A=\left(\frac{10x-5}{x^2-9}-\frac{3-x}{x+3}\right):\frac{x+2}{x+3}\)
\(=\left(\frac{10x-5}{\left(x-3\right)\left(x+3\right)}+\frac{x-3}{x+3}\right):\frac{x+2}{x+3}\)
\(=\frac{10x-5+\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}=\frac{10x-5+x^2-6x+9}{\left(x-3\right)}\cdot\frac{1}{x+2}\)
\(=\frac{x^2+4x+4}{\left(x+2\right)\left(x-3\right)}=\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-3\right)}=\frac{x+2}{x-3}\)
b: Thay x=5/2 vào A, ta được:
\(A=\left(\frac52+2\right):\left(\frac52-3\right)=\frac92:\frac{-1}{2}=-9\)
c: Để A nguyên thì x+2⋮x-3
=>x-3+5⋮x-3
=>5⋮x-3
=>x-3∈{1;-1;5;-5}
=>x∈{4;2;8;-2]
Kết hợp ĐKXĐ, ta được: x∈{4;2;8}
