a) Phân thức A được xác định khi: \(x^2-1\ne0\Rightarrow\left(x-1\right)\left(x+1\right)\ne0\Rightarrow\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Vây ĐKXĐ của A là \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b)Ta có: \(A=\dfrac{x^2+2x+1}{x^2-1}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)}{\left(x-1\right)}\)
Vậy \(A=\dfrac{x+1}{x-1}\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
c) Ta có A=2 <-> \(\dfrac{x+1}{x-1}=2\Leftrightarrow x+1=2\left(x-1\right)\Leftrightarrow x+1=2x-2\)
\(\Leftrightarrow x+1-2x+2=0\Leftrightarrow3-x=0\Rightarrow x=3\)
Vậy khi x=3 thì A=2
