Theo đề bài ta có :
\(A=\frac{n+1}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow2\left(n+1\right)=n-1\)
\(\Leftrightarrow2n+2=n-1\)
\(\Leftrightarrow2n-n=-1-2\)
\(\Rightarrow n=-3\)
Vậy với n = - 3 thì A = \(\frac{1}{2}\)
ĐKXĐ: \(n\ne1\)
\(\frac{n+1}{n-1}=\frac{n-1+2}{n-1}=1+\frac{2}{n-1}\)
mà \(A=\frac{1}{2}\)
\(\Rightarrow\)\(1+\frac{2}{n-1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{2}{n-1}=-\frac{1}{2}\)
\(\Leftrightarrow n-1=-4\)
\(\Leftrightarrow n=-3\) (t/m ĐKXĐ)