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\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\left(x;y\in Z\right)\)
\(MSC:8x\left(x\ne0\right)\)
\(pt\Leftrightarrow\dfrac{40+2xy}{8x}=\dfrac{x}{8x}\)
\(\Leftrightarrow40+2xy=x\)
\(\Leftrightarrow x-2xy=40\)
\(\Leftrightarrow x\left(1-2y\right)=40\)
\(\Leftrightarrow x;\left(1-2y\right)\in U\left(40\right)=\left\{-1;1;-2;2;-4;4;-5;5;-8;8;-10;10;-20;20;-40;40\right\}\)
Bạn lập bảng sẽ tìm ra các cặp \(\left(x;y\in Z\right)\) nhé!
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a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
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\(\dfrac{4}{x}-\dfrac{y}{2}=\dfrac{1}{4}\Leftrightarrow\dfrac{8-xy}{2x}=\dfrac{1}{4}\Leftrightarrow\dfrac{16-2xy}{4x}=\dfrac{x}{4x}\)
\(\Rightarrow16-2xy=x\Leftrightarrow x+2xy=16\Leftrightarrow x\left(1+2y\right)=16\)
\(\Rightarrow x;1+2y\inƯ\left(16\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
x | 1 | -1 | 2 | -2 | 4 | -4 | 8 | -8 | 16 | -16 |
2y + 1 | 16 | -16 | 8 | -8 | 4 | -4 | 2 | -2 | 1 | -1 |
y | 15/2 ( ktm ) | -17/2 ( ktm ) | 7/2 ( ktm ) | -9/2 ( ktm ) | 3/2 ( ktm ) | -5/2 ( ktm ) | 1/2 ( ktm ) | -3 / 2 ( ktm ) | 0 | -1 |
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\(y=\dfrac{2\left(2x+5\right)-18}{2x+5}=2-\dfrac{18}{2x+5}\)
\(y\in Z\Rightarrow\dfrac{18}{2x+5}\in Z\Rightarrow2x+5=Ư\left(18\right)\)
Mà 2x+5 luôn lẻ nên ta có: \(2x+5=\left\{-9;-3;-1;1;3;9\right\}\)
2x+5 | -9 | -3 | -1 | 1 | 3 | 9 |
x | -7 | -4 | -3 | -2 | -1 | 2 |
y | 4 | 8 | 20 | -16 | -4 | 0 |
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c, x/2+1/y=1/3 (x,y∈Z)
⇒1/y=1/3-x/2
⇒1/y=2-3x/6
⇒y(2-3x)=6
⇒y∈Ư(6)∈{1;-1;2;-2;3;-3;6;-6}
y | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
2-3x | 6 | -6 | 3 | -3 | 2 | -2 | 1 | -1 |
3x | -4 | 8 | -1 | 5 | 0 | 4 | 1 | 3 |
x | -4/3 (loại) | 8/3(loại) | -1/3(loại) | 5/3(loại) | 0 | 4/3(loại) | 1/3(loại) | 1
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Vậy các cặp (x;y) thỏa mãn pt trên là (0;3);(1;-6)
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\(\Leftrightarrow\left\{{}\begin{matrix}2^{2x}=2^3\cdot2^{x+y}\\3^{2x+2y}=3^5\cdot3^{5y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=x+y+3\\2x+2y=5y+5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=3\\2x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
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3/ Ta có:
\(A=\dfrac{1-2x}{x+3}\)
\(A=\dfrac{-2x+1}{x+3}\)
\(A=\dfrac{-2x-6+7}{x+3}\)
\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)
\(A=-2+\dfrac{7}{x+3}\)
A nguyên khi \(\dfrac{7}{x+3}\) nguyên
⇒ 7 ⋮ \(x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)
ĐKXĐ: ...
\(\Leftrightarrow40+2xy=x\)
\(\Leftrightarrow x\left(1-2y\right)=40\)
Do 40 có đúng 2 ước lẻ là 1 và -1; \(1-2y\) lẻ nên ta có các trường hợp:
\(\left[{}\begin{matrix}1-2y=1\\x=40\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=0\\x=40\end{matrix}\right.\)
\(\left[{}\begin{matrix}1-2y=-1\\x=-40\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\\x=-40\end{matrix}\right.\)