a) \(\frac{x^3+2x^2+x}{x^3-x}\Leftrightarrow\frac{x^3+2x^2+x}{x\left(x^2-1\right)}\Leftrightarrow\frac{x^3+2x^2+x}{x\left(x+1\right)\left(x-1\right)}\)
=> \(x\ne\){0,-1,1}
b) \(\frac{x^3+2x^2+x}{x^3-x}\)
\(\Leftrightarrow\frac{x\left(x^2+2x+1\right)}{x\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)^2}{x\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\frac{x+1}{x-1}\)
c) \(\frac{x+1}{x-1}=2\)
\(\Leftrightarrow\frac{x-1+2}{x-1}=2\)
\(\Leftrightarrow\frac{x-1}{x-1}+\frac{2}{x-1}=2\)
\(\Leftrightarrow1+\frac{2}{x-1}=2\)
\(\Leftrightarrow\frac{2}{x-1}=1\)
<=> x-1=2
<=> x=1 (KTMĐK)
Vậy không có giá trị x thõa mãn
b, ĐKXĐ : \(x\ne0\)
\(\frac{x^3+3x^2+x}{x^3-x}\)
= \(\frac{x\left(x^2+2x+1\right)}{x\left(x^2-1\right)}\)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
= \(\frac{x+1}{x-1}\)
c, \(\frac{x+1}{x-1}=2\)
ĐKXĐ : \(x\ne1\)
=> \(x+1=2\left(x-1\right)\)
=> \(x+1=2x-2\)
=> \(x-2x=-2-1\)
=> \(x=3\)
Vậy \(x\in\left\{3\right\}\)
