ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{5\left(x-1\right)}{2x}\)
\(=\frac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{5\left(x-1\right)}{2x}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{x+1}\cdot\frac{5}{2x}\)
\(=\frac{4x\cdot5}{2x\left(x+1\right)}\)
\(=\frac{10}{x+1}\)
e) Để A<0 thì \(\frac{10}{x+1}< 0\)
\(\Leftrightarrow\)10 và x+1 khác dấu
mà 10>0
nên x+1<0
hay x<-1(thỏa mãn ĐKXĐ)
Vậy: Khi x<-1 thì A<0
f) Để A nhận giá trị nguyên thì \(10⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(10\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{0;-2;1;-3;4;-6;9;-11\right\}\)
mà \(x\notin\left\{1;-1\right\}\)
nên \(x\in\left\{0;-2;-3;4;-6;9;-11\right\}\)
Vậy: Khi \(x\in\left\{0;-2;-3;4;-6;9;-11\right\}\) thì A nhận giá trị nguyên
