a, Không rõ đề bạn ơi ;-;
b, ĐKXĐ : \(x\ge0\)
Ta có : \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x=\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)=\left(\sqrt{x}-2\right)\left(2+\sqrt{x}\right)\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-5-\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow x=4\) ( TM )
Vậy ...
`b)(sqrtx-2)(5-sqrtx)=4-x`
`đk:0<=x`
`pt<=>(sqrtx-2)(sqrtx-5)=x-4`
`<=>x-7sqrtx+10=x-4`
`<=>7sqrtx=14`
`<=>sqrtx=2`
`<=>x=4(tmđk).`
a) Ta có: \(2\sqrt{x+2}=\sqrt{x^3-8}\)
\(\Leftrightarrow2\sqrt{x+2}-\sqrt{x+2}\cdot\sqrt{x^2-2x+4}=0\)
\(\Leftrightarrow\sqrt{x+2}\left(2-\sqrt{x^2-2x+4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-2x+4=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\\x=2\end{matrix}\right.\)
Vậy: S={0;-2;2}
