\(x^2+x+6+2x\sqrt{x+3}=4\left(x+\sqrt{x+3}\right)\) (*)
\(ĐK:x+3\ge0\)
\(\Leftrightarrow x\ge-3\)
(*)\(\Leftrightarrow x^2+2x\sqrt{x+3}+x+3+3=4\left(x+\sqrt{x+3}\right)\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)^2+3=4\left(x+\sqrt{x+3}\right)\)
Đặt \(x+\sqrt{x+3}=a\)
\(\Leftrightarrow a^2+3=4a\)
\(\Leftrightarrow a^2-4a+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\) ( Vi-ét )
`@` Với `a=1`
\(\rightarrow x+\sqrt{x+3}=1\)
Đặt \(x+3=b;b\ge0\)
\(\Leftrightarrow a+\sqrt{a}=4\)
\(\Leftrightarrow a+\sqrt{a}-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{-1+\sqrt{17}}{2}\left(tm\right)\\a=\dfrac{-1-\sqrt{17}}{2}\left(ktm\right)\end{matrix}\right.\)
\(\rightarrow x+3=\dfrac{-1+\sqrt{17}}{2}\)
\(\Leftrightarrow x=\dfrac{-7+\sqrt{17}}{2}\left(tm\right)\)
`@`Với \(x+\sqrt{x+3}=3\)
Đặt \(x+3=c\)
\(\Leftrightarrow a+\sqrt{a}=6\)
\(\Leftrightarrow a+\sqrt{a}-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=-3\left(ktm\right)\end{matrix}\right.\)
\(\rightarrow x+3=2\)
\(\Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-7+\sqrt{17}}{2};-1\right\}\)
