a) Ta có :
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5}{6}-\dfrac{2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\dfrac{5-2y}{6}=\dfrac{4}{x}\)
\(\Leftrightarrow\left(5-2y\right)x=6.4=24\)
Vì \(x,y\in N\Leftrightarrow5-2y\in N;5-2y;x\inƯ\left(24\right)\)
Ta có bảng :
| \(x\) | \(y\) | \(5-2y\) | \(Đk\) \(x,y\in N\) |
| \(1\) | \(\dfrac{-19}{2}\) | \(24\) | loại |
| \(2\) | \(\dfrac{-7}{2}\) | \(12\) | loại |
| \(3\) | \(\dfrac{-3}{2}\) | 2\(8\) | loại |
| \(4\) | \(\dfrac{1}{2}\) | \(6\) | loại |
| \(8\) | \(1\) | \(3\) | thỏa mãn |
| \(12\) | \(\dfrac{3}{2}\) | \(2\) | loại |
| \(24\) | \(2\) | \(1\) | thỏa mãn |
Vậy ...
\(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{2y}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)
\(\Rightarrow x\left(5-2y\right)=24\)
\(\Rightarrow x;5-2y\inƯ\left(24\right)\)
Xét ước là xong
\(3x-xy-4y+12=17\)
\(\Rightarrow x\left(3-y\right)+4\left(3-y\right)=17\)
\(\Rightarrow\left(x+4\right)\left(3-y\right)=17\)
\(\Rightarrow x+4;3-y\inƯ\left(17\right)\)
\(Ư\left(17\right)=\left\{\pm1;\pm17\right\}\)
Xét ước
