a) ĐKXĐ: x≠-1; x≠1
Ta có: \(A=\left(\frac{3}{x+1}-\frac{1}{1-x}-\frac{8}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{3}{x+1}+\frac{1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\left(\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{\left(1-2x\right)}\)
\(=\frac{3x-3+x+1+8}{\left(x+1\right)\left(x-1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\frac{4x+6}{1-2x}\)
b) Để A có giá trị nguyên thì 4x+6⋮1-2x
hay 6-12x+16x⋮1-2x
⇔16x⋮1-2x
⇔16x-8+8⋮1-2x
⇔8⋮1-2x
⇔1-2x∈Ư(8)
⇔1-2x∈{1;-1;2;-2;4;-4;8;-8}
⇔2x∈{0;2;-1;3;-3;-5;-7;9}
hay \(x\in\left\{0;1;\frac{-1}{2};\frac{3}{2};-\frac{3}{2};-\frac{5}{2};-\frac{7}{2};-\frac{9}{2}\right\}\)
mà x∈Z và x≠\(\pm\)1
nên x=0
Vậy: Khi x=0 thì A có giá trị nguyên