ĐKXĐ: ...
\(\Leftrightarrow2\left(x+1\right)^2=9x\left(\frac{x+1}{\sqrt{x+2}+1}\right)^2\)
TH1: \(\left(x+1\right)^2=0\)
TH2: \(2=\frac{9x}{\left(\sqrt{x+2}+1\right)^2}\)
\(\Leftrightarrow2\left(x+3+2\sqrt{x+2}\right)=9x\)
\(\Leftrightarrow4\sqrt{x+2}=7x-6\) (\(x\ge\frac{6}{7}\))
\(\Leftrightarrow16\left(x+2\right)=\left(7x-6\right)^2\)
\(\Leftrightarrow...\)