`1/a+1/b+1/c=4/5`

Giả sử $a≥b≥c>0$ thì `a,b,cin N` do `1/a+1/b+1/c=4/5<1`. 

Vậy `a,b,c>1` và `1/a≤1/b≤1/c`.

`=>1/a+1/b+1/c≤1/c+1/c+1/c=3/c`

`=>4/5≤3/c`

`=>4c≤15`

`=>c≤3 3/4`

`=>c in{1;2;3}`

Nếu \(\left[ \begin{array}{l}c=1\\c=2\\c=3\end{array} \right.\)`=>` thì \(\left[ \begin{array}{l}\dfrac{1}{c}=1\\\dfrac{1}{c}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{3}\end{array} \right.\)

`=>` \(\left[ \begin{array}{l}\dfrac{1}{a}+\dfrac{1}{b}+1=\dfrac{4}{5}-1=\dfrac{-1}{5}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{2}=\dfrac{4}{5}-\dfrac{1}{2}=\dfrac{3}{10}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{3}=\dfrac{4}{5}-\dfrac{1}{3}=\dfrac{7}{15}\end{array} \right.\) 

`=>` Ta có `1/a≤1/b`

`=>1/a+1/b≤1/b+1/b=2/b`

`=>3/10≤2/b`

\(\left[ \begin{array}{l}\\\dfrac{1}{a}+\dfrac{1}{b}\leq\dfrac{1}{b}+\dfrac{1}{b}=\dfrac{2}{b}\\\left\{ \begin{array}{l}\\{\dfrac{3}{10}\leq\dfrac{2}{b}\\\\3b\leq 20\\b\leq 6 \dfrac{2}{3}}\\b\in\{1;2;3;4;5;6\}\\\\{\dfrac{7}{15}\leq \dfrac{2}{b}}\\7b\leq30\\b\leq4\dfrac{1}{15}\\b\in\{1;2;3;4\}\end{array}\right.\end{array} \right.\)

Ta có các trường hợp:

\(\left[ \begin{array}{l}\left\{\begin{array}{l}\\\dfrac{1}{a}+1+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{-7}{10}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{-1}{5}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{-1}{30}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{1}{20}⇒a=20\\\dfrac{1}{a}+\dfrac{1}{5}+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{1}{10}⇒a=10\\\dfrac{1}{a}+\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{2}{15}∉N(ktm)\end{array} \right.\\\left\{ \begin{array}{l}\dfrac{1}{a}+1+\dfrac{1}{3}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{-8}{15}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{4}{5}⇒\dfrac{-1}{30}<0(ktm)\\\dfrac{1}{a}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{2}{15}∉N(ktm)\\\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{3}=\dfrac{4}{5}⇒\dfrac{1}{a}=\dfrac{13}{60}∉N(ktm)\end{array} \right.\end{array} \right.\)

`1/a+1/2+1/4=4/5` với `a=20`

`1/a+1/5+1/2=4/5` với `a=10`

Vậy ta có các cặp `a,b,c` thõa mãn điều kiện là `(2,4,20);(2,5,10)`.