Question

Cho A= \(\frac{x}{x+1}+\frac{2x}{x^2-1}-\frac{1}{1-x}\)

 tìm ĐKXĐ và rút gọn A

 

Answer 1

\(A=\frac{x}{x+1}+\frac{2x}{x^2-1}-\frac{1}{1-x}=\frac{x}{x+1}+\frac{2x}{\left(x+1\right)\left(x-1\right)}+\frac{1}{-\left(1-x\right)}.\)
\(=\frac{x}{x+1}+\frac{2x}{\left(x+1\right)\left(x-1\right)}+\frac{1}{x-1}\)
\(=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{2x}{\left(x+1\right)\left(x-1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x\left(x-1\right)+2x+1\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-x+2x+1x+1}{\left(x+1\right)\left(x-1\right)}=\frac{x^2+2x+1}{\left(x+1\right)\left(x-1\right)}.\)
\(=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{x+1}{x-1}\)

Answer 2

ĐKXĐ: \(\hept{\begin{cases}x+1\ne0\Rightarrow x\ne-1\\1-x\ne0\Rightarrow x\ne1\end{cases}}\)
Vậy đkxđ là : \(x\ne+1,-1\)