ĐKXĐ: \(\left\{{}\begin{matrix}u\ne\frac{1}{3}\\u\ne-\frac{11}{3}\end{matrix}\right.\)
\(\frac{1}{\left(3u-1\right)^2}-\frac{3}{\left(3u+11\right)^2}+\frac{2}{\left(3u-1\right)\left(3u+11\right)}=0\)
\(\Leftrightarrow\left(3u+11\right)^2-3\left(3u-1\right)^2+2\left(3u-1\right)\left(3u+11\right)=0\)
\(\Leftrightarrow\left(3u+11\right)^2-\left(3u-1\right)\left(3u+11\right)+3\left[\left(3u-1\right)\left(3u+11\right)-\left(3u-1\right)^2\right]=0\)
\(\Leftrightarrow12\left(3u+11\right)-36\left(3u-1\right)=0\)
\(\Leftrightarrow3u=7\Rightarrow u=\frac{7}{3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}1-3u\ne0\\3u+11\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3u\ne1\\3u\ne-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u\ne\frac{1}{3}\\u\ne-\frac{11}{3}\end{matrix}\right.\)
Ta có: \(\frac{2}{\left(1-3u\right)\left(3u+11\right)}=\frac{1}{9u^2-6u+1}-\frac{3}{\left(3u+11\right)^2}\)
\(\Leftrightarrow\frac{2}{\left(1-3u\right)\left(3u+11\right)}-\frac{1}{\left(3u-1\right)^2}+\frac{3}{\left(3u+11\right)^2}=0\)
\(\Leftrightarrow\frac{2\cdot\left(1-3u\right)\cdot\left(3u+11\right)}{\left(1-3u\right)^2\left(3u+11\right)^2}-\frac{\left(3u+11\right)^2}{\left(1-3u\right)^2\left(3u+11\right)^2}+\frac{\left(1-3u\right)^2\cdot3}{\left(3u+11\right)^2\left(1-3u\right)^2}=0\)
\(\Leftrightarrow\left(2-6u\right)\left(3u+11\right)-\left(9u^2+66u+121\right)+\left(1-6u+9u^2\right)\cdot3=0\)
\(\Leftrightarrow6u+22-18u^2-66u-9u^2-66u-121+3-18u+27u^2=0\)
\(\Leftrightarrow-144u-96=0\)
\(\Leftrightarrow-144u=96\)
\(\Leftrightarrow u=-\frac{96}{144}=-\frac{2}{3}\)(thỏa mãn)
Vậy: \(u=-\frac{2}{3}\)
