\(\dfrac{1}{a}+\dfrac{1}{b}=1\)
\(\Leftrightarrow a+b=ab\)(*)
Xét
\(\sqrt{a+b}=\sqrt{a-1}+\sqrt{b-1}\)
\(\Leftrightarrow a+b=a+b-2+2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow2=2\sqrt{ab-a-b+1}\)
\(\Leftrightarrow4=4\left(ab-a-b+1\right)\)
\(\Leftrightarrow4\left(ab-a-b\right)=0\Leftrightarrow ab-a-b=0\)
\(\Leftrightarrow ab=a+b\)(đúng với *)
\(\Rightarrow\)đpcm
Ta có\(\dfrac{1}{a}+\dfrac{1}{b}=1\Leftrightarrow\dfrac{a+b}{ab}=1\Leftrightarrow a+b=ab\Leftrightarrow ab-a-b=0\Leftrightarrow1=ab-a-b+1\Leftrightarrow1=a\left(b-1\right)-\left(b+1\right)\Leftrightarrow1=\left(a-1\right)\left(b-1\right)\Leftrightarrow1=\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow2=2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow0=-2+2\sqrt{\left(a-1\right)\left(b-1\right)}\Leftrightarrow a+b=a-1+2\sqrt{\left(a-1\right)}\sqrt{\left(b-1\right)}+b-1\Leftrightarrow a+b=\left(\sqrt{a+1}+\sqrt{b+1}\right)^2\Leftrightarrow\sqrt{a+b}=\sqrt{\left(\sqrt{a+1}+\sqrt{b+1}\right)^2}\Leftrightarrow\sqrt{a+b}=\sqrt{a+1}+\sqrt{b+1}\left(đpcm\right)\)
