ĐKXĐ:x\(\ge1\)
Ta có: \(\sqrt{x-1}+\sqrt{2}=\sqrt{x}+1\)<=> \(\left(\sqrt{x-1}+\sqrt{2}\right)^2=\left(\sqrt{x}+1\right)^2\)
<=> \(x-1+2\sqrt{2x-2}+2=x+2\sqrt{x}+1\)
<=> \(2\sqrt{2x-2}=2\sqrt{x}=>\sqrt{2x-2}=\sqrt{x}\)
<=> 2x-2=x
=> x=2
\(\sqrt{x-1}-1=\sqrt{x}-\sqrt{2}\)
\(\frac{x-2}{\sqrt{x-1}+1}=\frac{x-2}{\sqrt{x}+\sqrt{2}}\)
tự giải tiếp nha.