ĐKXĐ: x≠1; x≠0
Ta có: \(A=\frac{x+1}{x-1}+\frac{1-3x}{x^3+x}:\frac{x-1}{x^2+1}\)
\(=\frac{x+1}{x-1}+\frac{1-3x}{x\left(x^2+1\right)}\cdot\frac{x^2+1}{x-1}\)
\(=\frac{x+1}{x-1}+\frac{\left(1-3x\right)\cdot\left(x^2+1\right)}{x\cdot\left(x^2+1\right)\cdot\left(x-1\right)}\)
\(=\frac{x+1}{x-1}+\frac{1-3x}{x\left(x-1\right)}\)
\(=\frac{x\left(x+1\right)}{x\left(x-1\right)}+\frac{1-3x}{x\left(x-1\right)}\)
\(=\frac{x^2+x+1-3x}{x\left(x-1\right)}=\frac{x^2-2x+1}{x\left(x-1\right)}=\frac{\left(x-1\right)^2}{x\left(x-1\right)}=\frac{x-1}{x}\)
Để A nhận giá trị nguyên thì
\(x-1⋮x\)
\(\Leftrightarrow-1⋮x\)
\(\Leftrightarrow x\inƯ\left(-1\right)\)
\(\Leftrightarrow x\in\left\{1;-1\right\}\)
mà x≠1 và x≠0
nên x=-1
Vậy: Khi x=-1 thì A nhận giá trị nguyên
