d: ĐKXĐ: x>=1
\(\sqrt{x-1}-\sqrt{x+1}=2\)
=>\(\left(\sqrt{x-1}-\sqrt{x+1}\right)^2=2^2\)
=>\(x-1+x+1-2\cdot\sqrt{\left(x-1\right)\left(x+1\right)}=4\)
=>\(2\cdot\sqrt{x^2-1}=2x-4=2\left(x-2\right)\)
=>\(\sqrt{x^2-1}=x-2\)
=>\(\begin{cases}x-2\ge0\\ \left(x-2\right)^2=x^2-1\end{cases}\Rightarrow\begin{cases}x\ge2\\ x^2-4x+4-x^2+1=0\end{cases}\)
=>\(\begin{cases}x\ge2\\ -4x+5=0\end{cases}\Rightarrow x\in\) ∅
e: ĐKXĐ: x>=1
\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
=>\(\sqrt{x-1-2\cdot\sqrt{x-1}+1}-\sqrt{x-1}=1\)
=>\(\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{x-1}=1\)
=>\(\left|\sqrt{x-1}-1\right|-\sqrt{x-1}=1\) (1)
TH1: \(\sqrt{x-1}-1\ge0\)
=>\(\sqrt{x-1}\ge1\)
=>x-1>=1
=>x>=2
(1) sẽ trở thành: \(\sqrt{x-1}-1-\sqrt{x-1}=1\)
=>-1=1(vô lý)
TH2: 1<=x<=2
(1) sẽ trở thành: \(1-\sqrt{x-1}-\sqrt{x-1}=1\)
=>\(-2\cdot\sqrt{x-1}=0\)
=>\(\sqrt{x-1}=0\)
=>x-1=0
=>x=1(nhận)
