Có \(\frac{1}{a}+\frac{1}{b}=1\Rightarrow a+b=ab\)
Xét \(VP^2=\left(\sqrt{a-1}+\sqrt{b-1}\right)^2=a-1+b-1+2\sqrt{\left(a-1\right)\left(b-1\right)}\)
\(=a+b-2+2\sqrt{ab-a-b+1}=a+b-2+2\sqrt{a+b-a-b+1}=a+b=\left(\sqrt{a+b}\right)^2=VT^2\)
\(\Rightarrow\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)