a/ ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Vậy..
b/ Ta có :
\(C=\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)
\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{2x^2+2x+x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}\)
Vậy...
c/ Với mọi x ta có :
\(\left\{{}\begin{matrix}\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\\5>0\end{matrix}\right.\)
\(\Leftrightarrow\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}>0\)
\(\Leftrightarrow C>0\left(đpcm\right)\)