\(x+3=\sqrt{x-1}+2\sqrt{x+2}\)
\(\Leftrightarrow x+3-2\sqrt{x+2}=\sqrt{x-1}\)
\(\Leftrightarrow\left(\sqrt{x+2}-1\right)^2=\sqrt{x-1}\)
Đặt \(x+2=a^2\Rightarrow x-1=a^2-3\)
\(\left(a-1\right)^2=\sqrt{a^2-3}\)
\(\Leftrightarrow\left(a^2-2a+1\right)^2=a^2-3\)
\(\Leftrightarrow a^4+4a^2+1+2a^2-4a^3-4a=a^2-3\)
\(\Leftrightarrow a^4+5a^2-4a^3-4a+4=0\)
\(\Leftrightarrow\left(a^4-4a^3+4a^2\right)+a^2-4a+4=0\)
\(\Leftrightarrow\left(a^2-2a\right)^2+\left(a-2\right)^2=0\)
\(\Leftrightarrow a^2.\left(a-2\right)^2+\left(a-2\right)^2=0\)
\(\Leftrightarrow\left(a^2+1\right)\left(a-2\right)^2=0\)
\(\Leftrightarrow a-2=0\Leftrightarrow a=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=2\)
Vậy ...
Cách khác:
\(x\ge1\)
\(2x+6=2\sqrt{x-1}+4\sqrt{x+2}\)
\(\Leftrightarrow x+2-4\sqrt{x+2}+4+x-1-2\sqrt{x-1}+1=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-2\right)^2+\left(\sqrt{x-1}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+2}=2\\\sqrt{x-1}=1\end{matrix}\right.\) \(\Rightarrow x=2\)
