Giải:
Ta có: \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)
\(\Leftrightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}=\dfrac{3-n}{6}\)
\(\Leftrightarrow1.6=6=m\left(3-n\right)\)
Mà \(6=1.6=2.3=\left(-1\right).\left(-6\right)=\left(-2\right).\left(-3\right)\)
Ta có bảng sau:
| \(m\) | \(1\) | \(-1\) | \(6\) | \(-6\) | \(2\) | \(-2\) | \(3\) | \(-3\) |
| \(3-n\) | \(6\) | \(-6\) | \(1\) | \(-1\) | \(3\) | \(-3\) | \(2\) | \(-2\) |
| \(n\) | \(-3\) | \(9\) | \(2\) | \(4\) | \(0\) | \(6\) | \(1\) | \(5\) |
Vậy...
Ta có \(\dfrac{1}{m}+\dfrac{n}{6}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{1}{2}-\dfrac{n}{6}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{3}{6}-\dfrac{n}{6}\)
\(\Rightarrow\dfrac{1}{m}=\dfrac{3-n}{6}\)
\(\Rightarrow1\times6=\left(3-n\right)\times m\)
\(\Rightarrow6=\left(3-n\right)\times m\)
\(\Rightarrow\left(3-n\right);m\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow\left(3-n\right)\times m=6=(-1)\times\left(-6\right)=(-6)\times\left(-1\right)=\left(-2\right)\times\left(-3\right)=\left(-3\right)\times\left(-2\right)=1\times6=6\times1=2\times3=3\times2\)
Ta có bảng sau
| 3-n | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| m | -1 | -2 | -3 | -6 | 6 | 3 | 2 | 1 |
| n | 9 | 6 | 5 | 4 | 2 | 1 | 0 | -3 |
Vậy các cặp m,n thỏa mãn là
| m | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 |
| n | 9 | 6 | 5 | 4 | 2 | 1 | 0 | -3 |
