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NM
0
30 tháng 10 2017
x.x^2+6
x^2.2+6
x^4+6
x.x.x.x+6
con lai ban tu lam minh xin het
VM
0
TT
0
LM
2
AH
Akai Haruma
Giáo viên
19 tháng 7 2024
1/
Vì $ƯCLN(x,y)=6$ nên đặt $x=6m, y=6n$ với $m,n$ là số tự nhiên, $m,n$ nguyên tố cùng nhau.
Theo bài ra ta có:
$xy=720$
$\Rightarrow 6m.6n=720$
$\Rightarrow mn=20$
Do $m,n$ nguyên tố cùng nhau nên $(m,n)=(1,20), (4,5), (5,4), (20,1)$
$\Rightarrow (x,y)=(6,120), (24,30), (30,24), (120,60)$
AH
Akai Haruma
Giáo viên
19 tháng 7 2024
2/
Vì $5x=|x+2|+|2x+1|+|x+3|\geq 0$ nên $x\geq 0$
$\Rightarrow |x+2|=x+2; |2x+1|=2x+1; |x+3|=x+3$. Bài toán trở thành:
$x+2+2x+1+x+3=5x$
$\Rightarrow 4x+6=5x$
$\Rightarrow x=6$ (thỏa mãn)
NM
1
XO
21 tháng 1 2020
Ta có : xy + 5x - 2y = 13
=> x(y + 5) - 2y = 13
=> x(y + 5) - 2y - 10 = 13 - 10
=> x(y + 5) - 2(y + 5) = 3
=> (x - 2)(y + 5) = 3
Với \(x;y\inℤ\Rightarrow\hept{\begin{cases}x-2\inℤ\\y+5\inℤ\end{cases}}\)
mà 3 = 1.3 = (-1) . (-3)
Lập bảng xét các trường hợp
| x - 2 | 1 | 3 | -1 | -3 |
| y + 5 | 3 | 1 | -3 | -1 |
| x | 3 | 5 | 1 | -1 |
| y | -2 | -4 | -8 | -6 |
Vậy các cặp (x ; y) thỏa mãn là : (3 ; -2) ; (5 ; -4) ; (1; - 8) ; (-1;-6)
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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