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a, \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-3,2y-6\in Z\\x-3,2y-6\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\end{matrix}\right.\)
Ta có bảng:
| x-3 | -1 | -5 | 1 | 5 |
| 2y-6 | -5 | -1 | 5 | 1 |
| x | 2 | -2 | 4 | 8 |
| y | \(\dfrac{1}{2}\left(loại\right)\) | \(\dfrac{5}{2}\left(loại\right)\) | \(\dfrac{11}{2}\left(loại\right)\) | \(\dfrac{7}{2}\left(loại\right)\) |
Vậy không có x,y thỏa mãn đề bài
b, tương tự câu a
\(c,xy-5x+2y=7\\ \Rightarrow x\left(y-5\right)+2y-10=-3\\ \Rightarrow x\left(y-5\right)+2\left(y-5\right)=-3\\ \Rightarrow\left(x+2\right)\left(y-5\right)=-3\)
Rồi làm tương tự câu a
\(d,xy-3x-4y=5\\ \Rightarrow x\left(y-3\right)-4y+12=17\\ \Rightarrow x\left(y-3\right)-4\left(y-3\right)=17\\ \Rightarrow\left(x-4\right)\left(y-3\right)=17\)
Rồi làm tương tự câu a
a: =>x-xy+y=0
=>x(1-y)+1-y-1=0
=>(x+1)(1-y)=1
=>(x+1)(y-1)=-1
=>\(\left(x+1;y-1\right)\in\left\{\left(-1;1\right);\left(1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-2;2\right);\left(0;0\right)\right\}\)
b: 2x-xy-2y=3
=>x(2-y)-2y+4=7
=>x(2-y)+2(2-y)=7
=>(x+2)(y-2)=-7
=>\(\left(x+2;y-2\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(-1;-5\right);\left(-9;3\right);\left(-3;9\right);\left(5;1\right)\right\}\)
c: =>x(4-y)+5y-20=-3
=>x(4-y)-5(4-y)=-3
=>(4-y)(x-5)=-3
=>(x-5)(y-4)=3
=>\(\left(x-5;y-4\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(6;9\right);\left(8;5\right);\left(4;1\right);\left(2;3\right)\right\}\)
a, \(xy\) + 4\(x\) + \(y\) = 6
\(xy\) + y + 4\(x\) + 4 = 10
(\(xy\)+y) + (4\(x\) + 4) = 10
y(\(x\) + 1) + 44(\(x\) + 1) =10
(\(x\) + 1)(y + 4) = 10
Ư(10) = { -10; -5; -2; -1; 1; 2; 5; 10}
Lập bảng ta có:
| \(x+1\) | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| \(x\) | -11 | -6 | -3 | -2 | 0 | 1 | 4 | 9 |
| y + 4 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
| y | -5 | -6 | -9 | -14 | 6 | 1 | -2 | -3 |
Từ bảng trên ta có các cặp \(x\) , y nguyên thỏa mãn đề bài là:
(\(x\); y) =(-11; -5); ( -6; -6); (-3; -9); (-2; -14); (0; 6); (1; 1); (4; -2); (9; - 3)
b, \(xy\) - 2\(x\) = y - 3
\(x\)y - y - 2\(x\) + 2 = -1
(\(x\)y - y) - (2\(x\) - 2) = -1
y(\(x\) - 1) - 2(\(x\) -1) = -1
(\(x\) - 1)(y -2) = -1
⇔ (1-\(x\))(y-2) =1
Ư(1) = {-1; 1}
Lập bảng ta có:
| \(1-x\) | -1 | 1 |
| \(x\) | 2 | 0 |
| y- 2 | -1 | 1 |
| y | 1 | 3 |
Theo bảng trên ta có các cặp \(x\), y nguyên thỏa mãn đề bài là:
(\(x\); y) = (2; 1); (0; 3)
a) x + y +xy = 6
y( 1 + x ) + x + 1 = 7
( x + 1 ) ( y + 1 ) = 7
| x+1 | -7 | -1 | 1 | 7 |
| y+1 | -1 | -7 | 7 | 1 |
| x | -8 | -2 | 0 | 6 |
| y | -2 | -8 | 6 | 0 |
b) 2x + y - 2xy - 8 = 0
2x ( 1 - y ) - ( 1 - y ) - 7 = 0
( 1 - y ) ( 2x - 1 ) = 7
| 2x - 1 | -7 | -1 | 1 | 7 |
| 1 - y | -1 | -7 | 7 | 1 |
| x | -3 | 0 | 1 | 4 |
| y | 2 | 8 | -6 | 0 |
c) x - 4y + xy - 1 = 0
x( 1 + y ) -4( 1 + y ) + 3 = 0
( 1 + y ) ( x- 4 ) = 3
| x- 4 | -3 | -1 | 1 | 3 |
| 1 + y | -1 | -3 | 3 | 1 |
| x | 1 | 3 | 5 | 7 |
| y | -2 | -4 | 2 | 0 |
4:
(x+1)(y-2)=5
=>\(\left(x+1;y-2\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;7\right);\left(4;3\right);\left(-2;-3\right);\left(-6;1\right)\right\}\)
a, 3x ( y+1) + y + 1 = 7
(y+1)(3x +1) =7
th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)
th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)
th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)
Vậy (x,y)= (2 ;0); (0; 6)
b, xy - x + 3y - 3 = 5
(x( y-1) + 3( y-1) = 5
(y-1)(x+3) = 5
th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)
th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)
th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)
th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)
vậy (x, y) = ( 8; 2); ( -8; 0); (-2; 6); (-4; -4)
c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1
⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1 ⋮ 2x + 1
th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8
th2: 2x+ 1 = 1=> x =0; y = 7
th3: 2x+1 = -3 => x = x=-2 => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3
th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2
th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2
th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1
th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1
th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0
kết luận
(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)
3xy−2x+5y=293xy−2x+5y=29
9xy−6x+15y=879xy−6x+15y=87
(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77
3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77
(3y−2)(3x+5)=77(3y−2)(3x+5)=77
⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77
Ta có bảng giá trị sau:
Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}
Bài 1: Ta có 5x+7=5(x-2)+8
Để 5x+7 chia hết cho x-2 thì 5(x-2) +8 chia hết cho x-2
=> 8 chia hết cho x-2
x nguyên => x-2 nguyên => x-2 thuộc Ư (8)={-8;-4;-2;-1;1;2;4;8}
ta có bảng
Bài 2:
a) xy+x=-15
<=> x(y+1)=-15
=> x, y+1 thuộc Ư (-15)={-15;-5;-3;-1;1;3;5;15}
Ta có bảng
b) xy+2-y=9
<=> y(x-1)=7
=> y, x-1 thuộc Ư (7)={-7;-1;1;7}
Ta có bảng
c) xy+2x+2y=-17
<=> x(y+2)+2(y+2)=-15
<=> (x+2)(y+2)=-15
<=> x+2; y+2 thuộc Ư (-15)={-15;-5;-3;-1;1;3;5;15}
Ta có bảng