a) \(đk:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
b) \(đk:x\ge0\)
\(\Leftrightarrow2\sqrt{x}\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
c) \(đk:x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(tm\right)\)
d) \(đk:x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-3\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow x=4\left(tm\right)\)
e) \(đk:x\ge0\)
\(\Leftrightarrow\left(2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-\dfrac{1}{2}\left(VLý\right)\\\sqrt{x}=-2\left(VLý\right)\end{matrix}\right.\)
Vậy \(S=\varnothing\)
