\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a-b}\)
\(\Rightarrow\dfrac{b\left(a-b\right)}{ab\left(a-b\right)}+\dfrac{a\left(a-b\right)}{ab\left(a-b\right)}=\dfrac{ab}{ab\left(a-b\right)}\left(a,b\ne0;a\ne b;a,b>0\right)\)
\(\Rightarrow\left(a-b\right)\left(b-a\right)=ab\)
\(\Rightarrow-\left(a-b\right)\left(a-b\right)=ab\)
\(\Rightarrow-\left(a-b\right)^2=ab\left(1\right)\)
mà \(\left\{{}\begin{matrix}-\left(a-b\right)^2< 0\\ab>0\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\) vô lý
⇒ không có 2 số a≠b; a,b>0 thỏa đề bài