Đáp án:
Theo đề, ra ta có:
\[\dfrac{5}{6} =\dfrac{1}{a}+\dfrac{1}{b}(a;b\in N^*)\]
\[\to \dfrac{1}{a}<\dfrac{5}{6}\]
\[\to a>1,2\qquad (1)\]
Giả sử $a<b$
$\to \dfrac{1}{a}>\dfrac{1}{b}$
$\to \dfrac{1}{a}+\dfrac{1}{a}>\dfrac{1}{a}+\dfrac{1}{b}$
$\to \dfrac{2}{a}>\dfrac{5}{6}\to a<2,4\qquad (2)$
Từ $(1);(2)→a=2($ vì $a\in N*)$
$→b=3$
$\dfrac{5}{6}=\dfrac{1}{2}+\dfrac{1}{3}$