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![](https://rs.olm.vn/images/avt/0.png?1311)
\(3xy+y=4-x\\ \Leftrightarrow y\left(3x+1\right)+x=4\\ \Leftrightarrow3y\left(3x+1\right)+\left(3x+1\right)=13\\ \Leftrightarrow\left(3y+1\right)\left(3x+1\right)=13\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}3x+1,3y+1\in Z\\3x+1,3y+1\inƯ\left(13\right)\end{matrix}\right.\)
Ta có bảng:
3x+1 | -13 | -1 | 1 | 13 |
3y+1 | -1 | -13 | 13 | 1 |
x | \(-\dfrac{14}{3}\left(ktm\right)\) | \(-\dfrac{2}{3}\left(ktm\right)\) | 0 | 4 |
y | \(-\dfrac{2}{3}\left(ktm\right)\) | \(-\dfrac{14}{3}\left(ktm\right)\) | 4 | 0 |
Vậy \(\left(x,y\right)\in\left\{\left(0;4\right);\left(4;0\right)\right\}\)
⇔3(3xy+y)=3(4−x)⇔3(3xy+y)=3(4-x)
⇔9xy+3y=12−3x⇔9xy+3y=12-3x
⇔9xy+3y+3x=12⇔9xy+3y+3x=12
⇔9xy+3y+(3x+1)=12+1=13⇔9xy+3y+(3x+1)=12+1=13
⇔3y.(3x+1)+(3x+1)=13⇔3y.(3x+1)+(3x+1)=13
⇔(3x+1)(3y+1)=13⇔(3x+1)(3y+1)=13
→→ (3x+1)(3x+1) và (3y+1)∈Ư(13)(3y+1)∈Ư(13)
Xét từng cặp ta dc :
+)+) x = 0 ; y=4
+)+) x = -2/3 , y = -14/3
+)+) x = 4 ; y = 0
+)+) x=−143x=-143 ; y = -2/3
Mà x;y∈Zx;y∈Z
⇒⇒ ( x ; y ) = ( 4 ; 0 ) ; ( 0 ; 4 )
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$x+4=3xy+y$
$x+4=y(3x+1)$
$3x+12=y(3x+1)$
$(3x+1)+11=y(3x+1)$
$11=y(3x+1)-(3x+1)=(y-1)(3x+1)$
$\Rightarrow 11\vdots y-1$
$\Rightarrow y-1\in\left\{1; -1; 11; -11\right\}$
$\Rightarrow y\in\left\{2; 0; 12; -10\right\}$
Với $y=2$ thì $3x+1=11\Rightarrow x=\frac{10}{3}$ (loại)
Với $y=0$ thì $3x+1=-11\Rightaarrow x=-4$
Với $y=12$ thì $3x+1=1\Rightarrow x=0$
Với $y=-10$ thì $3x+1=-1\Rightarrow x=\frac{-2}{3}$ (loại)
\(x\) + 4 = 3\(x\)y + y
\(x\) + 4 = y( 3\(x\)+1)
3(\(x+4\)) = 3y( 3\(x\)+1)
3\(x\) + 12 = 3y(3\(x\) + 1)
(3\(x\) + 1) + 11 = 3y(3\(x\)+ 1)
3y(3\(x\) + 1) - (3\(x\) +1 ) = 11
(3\(x\) +1)(3y -1) = 11
Ư(11) = { -11; -1; 1; 11}
Lập bảng ta có:
\(3x+1\) | -11 | -1 | 1 | 11 |
3y-1 | -1 | -11 | 11 | 1 |
\(x\) | -4 | -2/3 | 0 | 10/3 |
y | 0 | -10/3 | 4 | 2/3 |
Vậy cặp số \(x\),y thỏa mãn đề bài là:
(\(x\),y) = ( -4; 0); ( 0; 4)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y\left(3x+1\right)=4-x\)
\(3y\left(3x+1\right)=12-3x\)
\(3y\left(3x+1\right)=-\left(3x+1\right)+13\)
\(\left(3x+1\right)\left(3y+1\right)=13\)
Đến đây tự giải nhá.
![](https://rs.olm.vn/images/avt/0.png?1311)
3xy + y=4-x
<=>9xy+3y=12-3x
<=>9xy+3y+3x+1=13
<=>3y.(3x+1)+(3x+1)=13
<=>(3x+1)(3y+1)=13
<=> *\(\begin{cases}3x+1=1\\3y+1=13\end{cases}\)<=>\(\begin{cases}x=0\\y=4\end{cases}\)(nhận)
*\(\begin{cases}3x+1=12\\3y+1=1\end{cases}\)<=>\(\begin{cases}x=4\\y=0\end{cases}\)(nhận)
*\(\begin{cases}3x+1=-1\\3y+1=-13\end{cases}\)<=>\(\begin{cases}x=-\frac{2}{3}\\y=-\frac{14}{3}\end{cases}\)(loại)
*\(\begin{cases}3x+1=-13\\3y+1=-1\end{cases}\)<=>\(\begin{cases}x=-\frac{14}{3}\\y=-\frac{2}{3}\end{cases}\)(loại)
Vậy x=4 thì y=0 ; x=0 thì y=4
![](https://rs.olm.vn/images/avt/0.png?1311)
e) \(x+y+3xy=1\)
\(\Leftrightarrow3x+3y+9xy=3\)
\(\Leftrightarrow3x+9xy+3y=3\)
\(\Leftrightarrow3x\left(1+3y\right)+1+3y=4\)
\(\Leftrightarrow\left(3y+1\right)\left(3x+1\right)=4\)
\(\Leftrightarrow\left(3x+1\right);\left(3y+1\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-\dfrac{2}{3};-\dfrac{5}{3}\right);\left(0;1\right);\left(-1;-1\right);\left(\dfrac{1}{3};\dfrac{1}{3}\right);\left(-\dfrac{5}{3};-\dfrac{2}{3}\right);\left(1;0\right)\right\}\)
\(\Leftrightarrow\left(x;y\right)\in\left\{\left(0;1\right);\left(-1;-1\right);\left(1;0\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3xy + y=4-x
<=>9xy+3y=12-3x
<=>9xy+3y+3x+1=13
<=>3y.(3x+1)+(3x+1)=13
<=>(3x+1)(3y+1)=13
<=> *{3x+1=13y+1=13{3x+1=13y+1=13<=>{x=0y=4{x=0y=4(nhận)
*{3x+1=123y+1=1{3x+1=123y+1=1<=>{x=4y=0{x=4y=0(nhận)
*{3x+1=−13y+1=−13{3x+1=−13y+1=−13<=>{x=−23y=−143{x=−23y=−143(loại)
*{3x+1=−133y+1=−1{3x+1=−133y+1=−1<=>{x=−143y=−23{x=−143y=−23(loại)
Vậy x=4 thì y=0 ; x=0 thì y=4
co ai biet nguyen trung nghia