đặt \(\left\{{}\begin{matrix}a=x+1\\b=x-2\end{matrix}\right.\)
pt\(\Leftrightarrow2a-5\sqrt{ab}+2b=0\) đk \(x\ge2\)
\(\Leftrightarrow\left(2a-4\sqrt{ab}\right)-\left(\sqrt{ab}-2b\right)=0\)
\(\Leftrightarrow2\sqrt{a}\left(\sqrt{a}-2\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-2\sqrt{b}\right)=0\)
\(\Leftrightarrow\left(2\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-2\sqrt{b}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{a}=\sqrt{b}\\\sqrt{a}=2\sqrt{b}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4(x+1)=x-2\\x+1=4\left(x-2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)
