\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x\inƯ\left(40\right);1-2y\inƯ\left(40\right)\)
mà \(1-2y\) lẻ nên \(1-2y\in\left\{\pm1;\pm5\right\}\)
Xét 4 trường hợp: +) Th1: Nếu \(1-2y=1\) thì \(x=40\)
\(\Rightarrow y=0\)
+) Th2: nếu \(1-2y=-1\) thì \(x=-40\)
\(\Rightarrow y=1\)
+) TH3: Nếu \(1-2y=5\) thì \(x=8\)
\(\Rightarrow y=-2\)
+) Th4: Nếu \(1-2y=-5\) thì \(x=-8\)
\(\Rightarrow y=3\)
Vậy ta tìm được các cặp số sau: \(\left(x,y\right)\in\left\{\left(40,0\right);\left(-40,1\right);\left(8,-2\right);\left(-8,3\right)\right\}\)
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}=\dfrac{1}{8}-\dfrac{2y}{8}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=5.8=40\)
Ta có 1 - 2y là ước lẻ của 40
\(\Rightarrow1-2y\in\left\{1,-1,5,-5\right\}\)
Ta có bảng sau:
| 1-2y | 1 | -1 | 5 | -5 |
| 2y | 0 | 2 | -4 | 6 |
| y | 0 | 1 | -2 | 3 |
* Nếu y = 0 \(\Rightarrow\) x = 40
* Nếu y = 1 \(\Rightarrow\) x = -40
* Nếu y = -2 \(\Rightarrow\) x = 8
* Nếu y = 3 \(\Rightarrow\) x = -8
- Ta có: 5/x +y/4 =1/8
=> (20+xy)/4x =1/8.
=> 8( 20 + xy ) = 4x
=> 2( 20 + xy ) = x
=> 40 + 2xy = x
=> 40 = x - 2xy
=> -40 = 2xy - x
=> 2xy - x = -40
=> x( 2y - 1 ) = -40
- Từ đây bạn tự giải nhé.
Ta có: \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}=\dfrac{1-2y}{8}\)
\(\Leftrightarrow x=\dfrac{8}{1-2y}.5\)
Dễ thấy \(\text{1-2y}\) là số lẻ nên \(\text{ ƯCLN(8;1-2y) = 1}\)\(\Rightarrow\dfrac{x}{8}=\dfrac{5}{1-2y}\)
mà \(\text{x, y}\) nguyên khi \(1-2y \) phải là ước của 5 <=> \(1 - 2y\) ∈ \(\left\{1;-1;5;-5\right\}\)
- Xét 1-2y = -1 => y = 1 => x = -40
- Xét 1-2y = 1 => y = 0 => x = 40
- Xét 1-2y = -5 => y = 3 => x = -8
- Xét 1-2y = 5 => y = -2 => x = 8
Vậy có 4 cặp (x,y) nguyên (-40;1) ; (40;0) ; (-8;-5) ; (8;5)
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}+\dfrac{2y}{8}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow40=x\left(1-2y\right)\)
\(\Rightarrow x;1-2y\inƯ\left(40\right)\)
\(Ư\left(40\right)=\left\{\pm1;\pm2;\pm5;\pm8;\pm20;\pm40\right\}\)
1-2y lẻ nên:
\(\Rightarrow\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=-2\Rightarrow y=-1\\x=-40\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)