NT
1
K
Khách
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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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V
5
1 tháng 3 2015
ta có :
xy-x+2y=3
xy-x+2y-3=0
xy-x+2y-3+1=1
x(y-1)+2(y-1)=1
(y-1)*(x+2)=1
=>(y-1) và (x+2) lần lượt là các cặp (1;1),(-1;-1)
Ta có y-1=1
<=>y=2
x+2=1
<=>x=-1
hoặc
y-1=-1
<=>0
x+2=-1
<=>x=-3
TN
1
TP
1
NX
0
NV
5
4 tháng 1 2020
Ta có: xy - x + 2y = 3
=> x(y - 1) + 2(y - 1) + 2 = 3
=> (x + 2)(y - 1) = 1
=> x + 2; y - 1 \(\in\)Ư(1) = {1; -1}
Lập bảng:
| x + 2 | 1 | -1 |
| y - 1 | 1 | -1 |
| x | -1 | -3 |
| y | 2 | 0 |
Vậy ....
L
4 tháng 1 2020
\(xy-x+2y=3\)
\(\Leftrightarrow x\left(y-1\right)+2y-2=1\)
\(\Leftrightarrow x\left(y-1\right)+2\left(y-1\right)=1\)
\(\Leftrightarrow\left(y-1\right)\left(x+2\right)=1\)
\(\Rightarrow y-1\) và \(x+2\) \(\inƯ\left(1\right)\)
\(\RightarrowƯ\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow\hept{\begin{cases}x+2=-1\\y-1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\y=2\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=1\\y-1=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=0\end{cases}}\)
Vậy \(\left(x;y\right)=\left(-3;2\right)\)
\(=\left(-1;0\right)\)
CM
0
ND
0
HM
1
xy-x+2y=3
=>x(y-1)+2y=3
=>x(y-1)+2y-2=1
=>x(y-1)+2(y-1)=1
=>(x+2)(y-1)=1
Ta có bảng:
x+2 -1 1
y-1 -1 1
x -3 -1
y 0 2
Vậy các cặp (x,y) thỏa mãn là:(-3,0);(-1,2)