Ta có \(\frac{1}{a}+\frac{1}{b}=1\)
⇔ \(\frac{a+b}{a\cdot b}=1\)
\(\Leftrightarrow a+b=a\cdot b\)
đặt \(A=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow A^2=\left(\sqrt{a-1}+\sqrt{b-1}\right)^2\)
\(\Leftrightarrow A^2=a+b-2+2\sqrt{\left(a-1\right)\cdot\left(b-1\right)}\)
\(\Leftrightarrow A^2=a+b-2+2\sqrt{a\cdot b-a-b+1}\)
\(\Leftrightarrow A^2=a+b-2+2\sqrt{a+b-a-b+1}\)
\(\Leftrightarrow A^2=a+b-2+2\sqrt{1}\)
\(\Leftrightarrow A^2=a+b-2+2\)
\(\Leftrightarrow A^2=a+b\)
\(\Leftrightarrow A=\sqrt{a+b}\)
\(\Leftrightarrow\sqrt{a-1}+\sqrt{b-1}=\sqrt{a+b}\) (đpcm)