a
ĐK: \(x\ge1\left(\sqrt{x-1}\ge0\right)\)
\(PT\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\\ \Leftrightarrow\left(\sqrt{x-1}\right)\left(\sqrt{x-2}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x-2}=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
b
ĐK: \(\left\{{}\begin{matrix}x^2-4x+4>0\\4x^2-4x+9>0\end{matrix}\right.\)
PT \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(2x-3\right)^2}\)
\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2x-3\\x-2=3-2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{5}{3}\left(nhận\right)\end{matrix}\right.\)