a, ĐKXĐ: \(X\ne0;X\ne\pm1\)
b,\(A=\frac{X\left(X^2+2X+1\right)}{X\left(X^2-1\right)}=\frac{X\left(X+1\right)^2}{X\left(X-1\right)\left(X+1\right)}=\frac{X+1}{X-1}\)
c,Ta có: \(A=\frac{X+1}{X-1}=2\Leftrightarrow2\left(X-1\right)=X+1\Leftrightarrow2X-2=X+1\Leftrightarrow X=3\)
a) \(ĐKXĐ:x\ne0;x\ne1\)
b) \(A=\frac{x^3+2x^2+x}{x^3-x}\)
\(A=\frac{x\left(x^2+2x+1\right)}{x\left(x^2-1\right)}\)
\(A=\frac{x\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)
\(A=\frac{x+1}{x-1}\)
vậy \(A=\frac{x+1}{x-1}\)
c) thay vào ta được \(\frac{x+1}{x-1}=2\)
\(\Rightarrow\left(x-1\right).2=x+1\)
\(\Rightarrow2x-2=x+1\)
\(\Rightarrow2x-x=1+2\)
\(\Rightarrow x=3\)
vậy \(x=3\)thì \(A=2\)