\(x\left(x^2-1\right)\sqrt{x-1}=0\)(ĐK:x\(\ge1\))
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\\\sqrt{x-1}=0\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\left(ktm\right)\\\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy S={1}
\(\sqrt{-x^2+6x-9}+x^3=27\Leftrightarrow\sqrt{-\left(x^2-6x+9\right)}+x^3=27\Leftrightarrow\sqrt{-\left(x-3\right)^2}+x^3=27\left(1\right)\)
Ta có \(\left(x-3\right)^2\ge0\Leftrightarrow-\left(x-3\right)^2\le0\)
Mà \(\sqrt{-\left(x-3\right)^2}\ge0\)
Suy ra \(\left\{{}\begin{matrix}-\left(x-3\right)^2=0\\x^3=27\end{matrix}\right.\)\(\Leftrightarrow x=3\left(tm\right)\)
Vậy S={3}
