Question

Tìm x, y,z nguyên dương sao cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

Answer 1

Không mất tính tổng quát, giả sử \(a\le b\le c\)

\(\Rightarrow\dfrac{1}{a}\ge\dfrac{1}{b}\ge\dfrac{1}{c}\)

\(\Rightarrow2=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}=\dfrac{3}{a}\)

\(\Rightarrow a\le\dfrac{3}{2}\)

Mà a là số nguyên dương

\(\Rightarrow a=1\)

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Rightarrow\dfrac{1}{b}+\dfrac{1}{c}=1\le\dfrac{1}{b}+\dfrac{1}{b}=\dfrac{2}{b}\)

\(\Rightarrow b\le2\)

\(\Rightarrow y\in\left\{1;2\right\}\)

\(\Rightarrow z\in\left\{1;2\right\}\)

Vậy \(\left(x;y;z\right)\in\left\{\left(1;2;2\right),\left(2;2;1\right),\left(2;1;2\right),\left(2;2;1\right)\right\}\)