a, ĐKXĐ:\(\left\{{}\begin{matrix}x^2-1\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm1\\x\ne-1\\x\ne1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)
b, \(P=\dfrac{2x^2}{x^2-1}+\dfrac{x}{x+1}-\dfrac{x}{x-1}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2+x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2+x^2-x-x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x^2-2x}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow P=\dfrac{2x}{x+1}\)
c, Thay x=2 vào P ta có:
\(P=\dfrac{2x}{x+1}=\dfrac{2.2}{2+1}=\dfrac{4}{3}\)
