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\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Rightarrow\dfrac{6}{6x}+\dfrac{2xy}{6x}=\dfrac{5x}{6x}\Rightarrow6+2xy=5x\)
\(\Rightarrow5x-2xy=6\Rightarrow x\left(5-2y\right)=6\)
Do \(x,y\) là số tự nhiên nên \(x\inƯ^+\left(6\right)\)
TH1: \(x=1\Rightarrow5-2y=6\Rightarrow y=-\dfrac{1}{2}\) (loại)
TH2: \(x=2\Rightarrow5-2y=3\Rightarrow y=1\) (TM)
TH3: \(x=3\Rightarrow5-2y=2\Rightarrow y=\dfrac{3}{2}\) (Loại)
TH4: \(x=6\Rightarrow5-2y=1\Rightarrow y=2\) (TM)
\(\Leftrightarrow6+2xy=5x\left(x\ne0\right)\)
\(\Leftrightarrow5x-2xy=6\Leftrightarrow x\left(5-2y\right)=6\)
\(\Leftrightarrow x=\dfrac{6}{5-2y}\)
Để x nguyên thì 5-2y phải là ước của 6
\(\Rightarrow5-2y=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow y=\left\{4;3;2;1\right\}\Rightarrow x=\left\{-2;-6;6;2\right\}\)
=>(12-xy)/3x=5/6
=>6(12-xy)=15x
=>(12-xy)=5/2x
=>24-2xy=5x
=>5x+2xy=24
=>x(2y+5)=24
=>(x;2y+5) thuộc {(1;24); (2;12); (3;8); (4;6); (6;4); (8;3); (12;2); (24;1)}
mà x,y là các số tự nhiên
nên \(\left(x,y\right)\in\varnothing\)
\(\dfrac{x}{9}=\dfrac{3}{y}+\dfrac{1}{18}\left(y\ne0\right)\)
\(\Rightarrow\dfrac{2xy}{18y}=\dfrac{54}{18y}+\dfrac{y}{18y}\)
\(\Rightarrow2xy=54+y\)
\(\Rightarrow2xy-y=54\)
\(\Rightarrow xy-\dfrac{y}{2}=27\)
\(\Rightarrow y\left(x-\dfrac{1}{2}\right)=27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right);y\in\left\{1;3;9;27\right\}\)
\(\Rightarrow\left(x;\right)y\in\left\{\left(\dfrac{1}{2};27\right);\left(\dfrac{5}{2};9\right);\left(\dfrac{17}{2};3\right);\left(\dfrac{53}{2};1\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\varnothing\left(x;y\inℕ\right)\)
Bài 2:
a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)
\(=\dfrac{4+6-3}{n-1}\)
\(=\dfrac{7}{n-1}\)
Để A là số tự nhiên thì \(7⋮n-1\)
\(\Leftrightarrow n-1\inƯ\left(7\right)\)
\(\Leftrightarrow n-1\in\left\{1;7\right\}\)
hay \(n\in\left\{2;8\right\}\)
Vậy: \(n\in\left\{2;8\right\}\)
ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2 Để B là STN thì 4n+10⋮n+2 4n+8+2⋮n+2 4n+8⋮n+2 ⇒2⋮n+2 n+2∈Ư(2) Ư(2)={1;2} Vậy n=0
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\)
\(\dfrac{x}{9}-\dfrac{1}{18}=\dfrac{3}{y}\)
\(\dfrac{2x}{18}-\dfrac{1}{18}=\dfrac{3}{y}\)
\(\dfrac{2x-1}{18}=\dfrac{3}{y}\)
\(\Rightarrow\)(2x-1).y=18.3=54
54 có các ước là: \(\pm1;\pm2;\pm3;\pm6;\pm9;\pm18;\pm27;\pm54\)
*\(\left\{{}\begin{matrix}2x-1=1\\y=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2\\y=54\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=54\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=54\\y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=55\\y=1\end{matrix}\right.\)\(\notin\) N ( Loại)
*\(\left\{{}\begin{matrix}2x-1=2\\y=27\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\y=27\end{matrix}\right.\) \(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=27\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=28\\y=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=14\\y=2\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=3\\y=18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=4\\y=18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=18\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}2x-1=18\\y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=19\\y=3\end{matrix}\right.\)\(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=6\\y=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=7\\y=9\end{matrix}\right.\)\(\notin\) N ( Loại )
*\(\left\{{}\begin{matrix}2x-1=9\\y=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=10\\y=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=6\end{matrix}\right.\)
Vậy có các cặp (x,y) t/m đề bài là : (1,54) ; (14,2) ; (2,18) ; (5,6)
bạn học trường nào?kết bạn nhé !