Tìm x,y nguyên biết x.(2y-1)2=32y+16
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OS
0
2 tháng 8 2020
a)Ta có:\(x-y=2\Rightarrow\left(x-y\right)^2=4\Rightarrow\left(x^2+y^2\right)-2xy=4\Rightarrow4-2xy=4\Rightarrow2xy=0\Rightarrow xy=0\)
Khi đó ta có:\(x^5y=xy^5=xy\left(x^4-y^4\right)=0\)
HL
2 tháng 8 2020
x2 - 5x - 2xy + 5y + y2 + 4
= (x2 - 2xy + y2) - (5x - 5y) + 4
= (x2 - xy - xy + y2) - 5.(x - y) + 4
= (x - y)2 - 5.1 + 4
= 1 - 5 + 4
= 0
DL
1
B
4
17 tháng 12 2021
Do 103 là số nguyên tố nên không chia hết cho 2
Mà 32y chia hết cho 2 nên \(5x^2⋮̸2\)
Mà 5 lẻ nên \(x^2\) lẻ
Do đó \(x^2\equiv1\left(mod4\right)\)
Lại có \(32y\equiv0\left(mod4\right)\Leftrightarrow5x^2-32y\equiv1\left(mod4\right)\)
Mà \(103\equiv3\left(mod4\right)\)
Vậy PT vô nghiệm
23 tháng 5 2016
a) \(xy+x+2y=5\Leftrightarrow xy+x+2y+2=7\Leftrightarrow\left(y+1\right)\left(x+2\right)=7\)
Vì x,y là số tự nhiên nên \(x,y\in N\)\(x,y\ge0\)\(\Rightarrow y+1\ge1;x+2\ge2\)
Từ đó ta có :
\(\hept{\begin{cases}x+2=7\\y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=0\end{cases}}}\)
b) \(xy+2x+2y=-16\Leftrightarrow xy+2y+2x+4=-12\Leftrightarrow\left(y+2\right)\left(x+2\right)=-12\)
Lần lượt xét từng trường hợp , ta được :
(x;y) = (-14; -1) ; (-8 ; 0) ; (-6 ; 1) ; (-5 ;2) ; (-4 ;4)
N
23 tháng 5 2016
a) \(\left(x+2\right)\left(y+1\right)=7=1.7=7.1\)
Hoặc \(\hept{\begin{cases}x+2=7\\y+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=0\end{cases}}}\in N\)
Hoặc\(\hept{\begin{cases}x+2=1\\y+1=7\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\notin N\\y=6\end{cases}}\)
Vậy \(\left(x;y\right)=\left(5;0\right)\)
b)\(\left(x+2\right)\left(y+2\right)=-1.12=-12.1=-2.6=-6.2=-3.4=-4.3\)
tương tự giải 6 TH là được
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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TD
0
HN
2
6 tháng 1 2019
(3x-1).y = -12<=> 3x-1 và y là Ư của -12 ={ 1;2;3;4;6;12∓1;2;3;4;6;12 }
=> ta xét từng trường hợp : ....
LL
6 tháng 1 2019
1. (3x-1)y=-12 suy ra \(3x-1\inƯ\left(-12\right)\)(em tự liệt kê nhé!)
Lại có 3x-1 chia 3 dư 2(thiếu 1) nên \(3x-1\in\left\{-1;2;-4;\right\}\)
Đến đây em lập bảng và tìm đáp số nhé!
2. \(5xy+5x+2y=-16\Rightarrow5x\left(y+1\right)+2y=-16\)
\(\Rightarrow5x\left(y+1\right)+2\left(y+1\right)=-16+2=-14\)
\(\Rightarrow\left(5x+2\right)\left(y+1\right)=14\)
\(\Rightarrow5x+2\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)mà 5x+2 lẻ nên \(5x+2\in\left\{\pm1;\pm7\right\}\)
Đến đây em hãy lập bảng và tìm ra đáp số nhé!
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