a) ĐKXĐ : \(x\ne1,-1,-\frac{1}{2}\)
\(A=\left(\frac{1}{x-1}-\frac{x}{1-x^3}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x+1}\right)\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Vậy : \(A=\frac{x+1}{x-1},\left(x\ne\pm1,-\frac{1}{2}\right)\)
b) Vì \(x=\frac{1}{2}\) thỏa mãn ĐKXĐ
Nên thay vào biểu thức A ta được :
\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=\frac{3}{2}:-\frac{1}{2}=\frac{3}{2}\cdot\left(-2\right)=-3\)
Vậy : \(A=-3\) với \(x=\frac{1}{2}\)
