\(\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-2\right)^2}=x-3\\ \Leftrightarrow\left|x-1\right|-\left|x-2\right|=x-3\left(1\right)\)
Ta có: \(x-1>x-2\) nên ta có 3 trường hợp:
`-` Xét trường hợp `1:`
\(\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
Khi đó:
\(\left(1\right)\Leftrightarrow x-1-\left(x-2\right)=x-3\\ \Leftrightarrow1=x-3\\ \Leftrightarrow x=4\left(TM\right)\)
`-` Xét trường hợp `2:`
\(\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=2-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x-2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\Leftrightarrow1\le x\le2\)
Khi đó:
\(\left(1\right)\Leftrightarrow x-1-\left(2-x\right)=x-3\\ \Leftrightarrow x-3=2x-3\\ \Leftrightarrow x=2x\\ \Leftrightarrow x=0\left(KTM\right)\)
`-` Xét trường hợp `3:`
\(\left\{{}\begin{matrix}\left|x-1\right|=1-x\\\left|x-2\right|=2-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-x\le0\\2-x\le0\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\le2\end{matrix}\right.\Leftrightarrow x\le1\)
Khi đó:
\(\left(1\right)\Leftrightarrow1-x-\left(2-x\right)=x-3\\ \Leftrightarrow3=x-3\\ \Leftrightarrow x=6\left(KTM\right)\)
Vậy `x=4`