a: \(\Leftrightarrow\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+2\right|=4\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}\)
Đặt \(\sqrt{x-1}=a\left(a>=0\right)\)
Theođề, ta có pt \(\left|a-1\right|=2-a\)
\(\Leftrightarrow\left\{{}\begin{matrix}a< =2\\\left(2-a-a+1\right)\left(2-a+a-1\right)=0\end{matrix}\right.\Leftrightarrow a=\dfrac{3}{2}\)
=>x-1=9/4
hay x=13/4
b: \(\Leftrightarrow\left|\sqrt{x-1}-3\right|+\left|\sqrt{x-1}-1\right|=4\)
Theo đề, ta có: \(\left|a-3\right|+\left|a-1\right|=4\)
TH1: 0<=a<1
Pt sẽ là -a+3-a+1=4
=>-2a=0
=>a=0
=>x=1
TH2: 1<=a<4
Pt sẽ là a-1+3-a=4
=>2=4(vô lý)
TH3: a>=4
Pt sẽ là a-3+a-1=4
=>2a=8
=>a=4
=>x-1=16
hay x=17
