Sửa đề: \(A=\left(\frac{2x}{x^2-9}-\frac{3}{x+3}-\frac{1}{3-x}\right):\frac{4}{x+3}\)
ĐKXĐ: x≠3; x≠-3
a) Ta có: \(A=\left(\frac{2x}{x^2-9}-\frac{3}{x+3}-\frac{1}{3-x}\right):\frac{4}{x+3}\)
\(=\left(\frac{2x}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x+3}{4}\)
\(=\frac{2x-3\left(x-3\right)+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}\)
\(=\frac{2x-3x+9+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{12}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{12\cdot\left(x+3\right)}{4\left(x+3\right)\left(x-3\right)}=\frac{3}{x-3}\)
b) Ta có: \(\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
mà x≠3; x≠-3
nên x=1
Thay x=1 vào biểu thức \(A=\frac{3}{x-3}\), ta được
\(\frac{3}{1-3}=-\frac{3}{2}\)
Vậy: \(\frac{-3}{2}\) là giá trị của biểu thức \(A=\left(\frac{2x}{x^2-9}-\frac{3}{x+3}-\frac{1}{3-x}\right):\frac{4}{x+3}\) tại x=1
c) Để A nhận giá trị nguyên thì \(3⋮x-3\)
⇔\(x-3\inƯ\left(3\right)\)
⇔\(x-3\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{4;2;6;0\right\}\)(tm)
Vậy: \(x\in\left\{4;2;6;0\right\}\)
