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VK
2
19 tháng 2 2018
Ta có : x2y - x + xy = 6
=> x(xy - 1 ) + xy = 6
=> x(xy-1)+xy-1=5
=>(xy-1)(x-1)=5
=>xy-1 ; x-1 thuộc Ư (5)
P/S: lập bảng là ok
ND
1
5 tháng 12 2016
\(xy\left(x+1\right)-x-1=5\)\(\Leftrightarrow xy\left(x+1\right)-\left(x+1\right)=5\)
\(\Leftrightarrow\left(x+1\right)\left(xy-1\right)=5=5.1=1.5\)số nguyễn thị thêm (-) nữa
\(\orbr{\begin{cases}x+1=1=>x=0\\xy-1=5=>\left(loai\right)\end{cases}}\)\(\hept{\begin{cases}x+1=5=>x=4\\4y-1=5=>y=\frac{6}{4}\left(loai\right)\end{cases}}\)
\(\hept{\begin{cases}x+1=-1=>x=-2\\-2y-1=-5=>y=2\left(nhan\right)\end{cases}}\)
\(\hept{\begin{cases}x+1=-5=>x=-6\\-6.y-1=-1=>y=0\end{cases}}\)
KL:
x,y=(-2,2)
x,y=(-6,0)
CM
0
LV
1
NH
1
11 tháng 5 2017
x2y - x +xy = 6
=> x[12y - 1 + y] = 6
=> xy = 6 [vì 12y - 1 = 0]
=> [x,y] = [1,6];[6,1];[-1,-6];[-6,-1];[2,3];[3,2];[-2,-3];[-3,-2]
Thử lại
* nếu x = 1; y = 6 thì x2y - x +xy = 6 [thỏa]
* nếu x = 6; y = 1 thì x2y - x +xy = 36 [loại]
* nếu x = -1; y = -6 thì x2y - x +xy = 8 [loại]
* nếu x = -6; y = -1 thì x2y - x +xy = 48 [loại]
* nếu x = 2; y = 3 thì x2y - x +xy = 68 [loại]
* nếu x = 3; y = 2 thì x2y - x +xy = 84 [loại]
* nếu x = -2; y = -3 thì x2y - x +xy = 513/64 [loại]
* nếu x = -3; y = -2 thì x2y - x +xy = 730/81 [loại]
Vậy [x,y] = [1;6]
LH
1
5 tháng 2 2020
a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng :
| x - 2 | -5 | -1 | 1 | 5 |
| y + 3 | -1 | -5 | 5 | 1 |
| x | -3 | 1 | 3 | 7 |
| y | -4 | -8 | 2 | -2 |
Vậy ......
b. Làm tương tự câu a.
c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6
d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1
14 tháng 1
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6 + xy = x + y
x + y - xy = 6
(x-1) + (y - xy) = 5
(x-1) - y.( x -1) = 5
(x-1)(1-y) = 5
Ư(5) = { -5; -1; 1; 5}
Lập bảng ta có :
Kết luận các cặp x, y nguyên thỏa mãn đề bài lần lượt là:
(x,y) = (-4; 2); ( 0; 6); ( 2; -4); ( 6; 0)
`6+xy=x+y`
`=>x+y-xy=6`
`=>x(1-y)-1+y=5`
`=>(x-1)(1-y)=5`
`@{(x-1=5),(1-y=1):}=>{(x=6),(y=0):}`
`@{(x-1=1),(1-y=5):}=>{(x=2),(y=-4):}`