ĐK: x\(\ge1\)
\(8\sqrt{x-1}-\sqrt{4x-4}+\sqrt{9x-9}=27\Leftrightarrow8\sqrt{x-1}-\sqrt{4\left(x-1\right)}+\sqrt{9\left(x-1\right)}=27\Leftrightarrow8\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=27\Leftrightarrow9\sqrt{x-1}=27\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x-1=9\Leftrightarrow x=10\left(tm\right)\)
Vậy S={10}
\(=>8\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=2\)
=> \(9\sqrt{x-1}=27\)
=> \(\sqrt{x-1}=3\)
=> \(\left\{{}\begin{matrix}3\ge0\\x-1=9< =>x=10\end{matrix}\right.\) luôn đúng
Vây, ...
\(\left(=\right)8\sqrt{x-1}-\sqrt{4\left(x-1\right)}+\sqrt{9\left(x-1\right)}=27\left(=\right)8\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=27\left(=\right)9\sqrt{x-1}=27\left(=\right)\sqrt{x-1}=27:9\left(=\right)\sqrt{x-1}=3\left(=\right)x-1=9\left(=\right)x=10\)
