\(1.\sqrt{x-1}=2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
\(2.\sqrt{3-x}=1\)
\(\Rightarrow3-x=1\)
\(\Rightarrow x=2\)
\(3.\left|x-1\right|+\left|x^2-1\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left|x^2-1\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(4.\left|2x-3\right|-\left|x-1\right|=0\)
\(\Rightarrow\left|2x-3\right|=\left|x-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=x-1\\2x-3=-x+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=3-1\\2x+x=3+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right..\)
