\(ĐK:4\le x\le6\)
\(PT\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}=6-x\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=6-x\)
\(\Leftrightarrow\sqrt{x-4}+2=6-x\)
\(\Leftrightarrow\sqrt{x-4}=4-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4=4-x\\4-x\ge0\end{matrix}\right.\)
=> x=4
Vậy.........
ĐKXĐ: \(x\ge4\)
Phương trình tương đương \(\sqrt{\left(\sqrt{x-4}+2\right)^2}=6-x\)
\(\left|\sqrt{x-4}+2\right|=6-x\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-4}=6-x-2=-x+4\\\sqrt{x-4}=x-6-2=x-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=\left(4-x\right)^2\forall x\le4\\x-4=\left(x-8\right)^2\forall x\ge8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\left(loai\right)\\x=\frac{17+\sqrt{17}}{2}\\x=\frac{17-\sqrt{17}}{2}\left(loai\right)\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4\\x=\frac{17+\sqrt{17}}{2}\end{matrix}\right.\)
