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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)
Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
Do đó \(3x+2\in\left\{-4;-1;2\right\}\)
\(\Rightarrow x\in\left\{-2;-1;0\right\}\)
Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)
Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)
Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)
\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)
Mà \(\left(5x-4\right)\left(y+3\right)=-18\)
\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)
Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)
Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)
Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)
Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)\left(y+1\right)=14\)
Do \(x,y\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2xy+y=7
=>(2x+1)*y=7
=>(2x+1;y) thuộc {(1;7); (7;1); (-1;-7); (-7;-1)}
=>(x,y) thuộc {(0;7); (3;1); (-1;-7); (-4;-1)}
b: =>(2x+1)^2+(y+1)^2=179-169=10
=>((2x+1)^2;(y+1)^2) thuộc {(1;9); (9;1)}
TH1: (2x+1)^2=1 và (y+1)^2=9
=>\(\left\{{}\begin{matrix}2x+1\in\left\{1;-1\right\}\\y+1\in\left\{3;-3\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{0;-1\right\}\\y\in\left\{2;-4\right\}\end{matrix}\right.\)
TH2: (2x+1)^2=9 và (y+1)^2=1
=>\(\left\{{}\begin{matrix}2x+1\in\left\{3;-3\right\}\\y+1\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{1;-2\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( Y + 1 ) X + Y + 1 = 10
<=> ( Y + 1 ) X + ( Y + 1 ) =10
<=> ( Y + 1 ) ( X + 1 ) = 10
X; Y thuộc Z nên X+1 ; Y +1 thuộc Z và \(\inƯ\left(10\right)\in\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Ta có bảng sau :
X+1 | -1 | -2 | -5 | -10 | 1 | 2 | 5 | 10 |
Y+1 | -10 | -5 | -2 | -1 | 10 | 5 | 2 | 1 |
X | -2 | -3 | -6 | -11 | 0 | 1 | 4 | 9 |
Y | -11 | -6 | -3 | -2 | 9 | 4 | 1 | 0 |
Vậy (X:Y) \(\in\){(-2;-11);(-3;-6);(-6;-3);(-11;-2);(0;9);(9;0);(1;4);(4;1)}
b) ( 2X +1)Y - 2X - 1 = -31
<=> ( 2X + 1)(Y-1) = -31
Vì X;Y \(\in\)Z
=> 2X+1 ; Y+1 \(\in\)Z
=> 2X+1 ; Y+1 \(\in\)Ư(-32)
Vì 2X là số chẵn với mọi X \(\in\)Z => 2X +1 là số lẻ với mọi X\(\in\)Z
Ta có bảng :
2X+1 | -1 | 1 |
Y-1 | 32 | -32 |
X | -1 | 0 |
Y | 33 | -31 |
Vậy ( X;Y ) \(\in\){ (-1;33);(0;-31)}
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: (x-2)(y+1)=-1
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+1=1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(3;-2);(1;0)}
b) Ta có: \(\left(2x+1\right)\left(y-2\right)=3\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1=1\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=3\\y-2=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=-1\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=-3\\y-2=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x=2\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}2x=-2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}2x=-4\\y=1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(0;5);(1;3);(-1;-1);(-2;1)}
thiẾu đầu bài?