c, ĐKXĐ: \(x\ne1\)
\(pt\Leftrightarrow\left(x-1\right)^2+\frac{9}{\left(x-1\right)^2}-7\left(x-1\right)+\frac{21}{x-1}=0\)
\(\Leftrightarrow\left(x-1-\frac{3}{x-1}\right)^2-7\left(x-1-\frac{3}{x-1}\right)+6=0\)
\(\Leftrightarrow\left(x+\frac{3}{x-1}-7\right)\left(x+\frac{3}{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{x-1}=7\\x+\frac{3}{x-1}=2\end{matrix}\right.\)
Nếu \(x+\frac{3}{x-1}=7\)
\(\Leftrightarrow x^2-8x+10=0\)
\(\Leftrightarrow x=4\pm\sqrt{6}\left(tm\right)\)
Nếu \(x+\frac{3}{x-1}=2\)
\(\Leftrightarrow x^2-3x+5=0\left(\text{vô nghiệm}\right)\)
Vậy pt fđã cho có hai nghiệm ...
