a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2-6x+9=2x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2-8x+12=0\end{matrix}\right.\Leftrightarrow x=6\)
b: \(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>x=0 hoặc x=4
`a)`\(\sqrt{2x-3}=x-3\)
\(ĐK:x\ge3\)
\(\Leftrightarrow\left(\sqrt{2x-3}\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow2x-3=x^2-6x+9\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{6\right\}\)
`b)`\(x-2\sqrt{x}=0\)
\(ĐK:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;4\right\}\)
