tìm số nguyên x thõa mãn:-210=-1+-2+-3+.......(-x+1)+-x
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NH
2
3 tháng 2 2016
a) (3x -2)^5 = -243
(3x -2)^5 = (-3)^5
3x-2=-3
3x=-1
x=-1/3
b) ( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
Do x phải là số nguyên tố nên x=7
HT
2
NN
2
LN
0
CT
3
MH
4 tháng 2 2016
-210 = (-1) + (-2) + (-3) + ... + (-x + 1) + (-x)
=> -210 = -(1 + 2 + 3 + ...+ x - 1 + x)
=> 210 = 1 + 2 + 3 + x - 1 + x
=> 210 = \(\frac{\left(x+1\right).x}{2}\)
=> 420 = (x + 1).x
=> 21 . 20 = (x + 1) . x
=> (20 + 1) . 20 = (x + 1) . x
=> x = 20
25 tháng 1 2017
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NT
4
TM
25 tháng 12 2016
Vì x;y nguyên nên (2x-3)2 và |y-2| đều là số nguyên
Mà \(\hept{\begin{cases}\left(2x-3\right)^2\ge0\\\left|y-2\right|\ge0\end{cases}}\) nên (2x-3)2 và |y-2| là các số nguyên không âm
TH1: (2x-3)2=0 và |y-2|=1
\(\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)(loại)
Ta không xét đến |y-2|=1 nữa!
TH2: (2x-3)2=1 và |y-2|=0
- \(\left(2x-3\right)^2=1\Rightarrow\orbr{\begin{cases}2x-3=-1\\2x-3=1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
- \(\left|y-2\right|=0\Leftrightarrow y-2=0\Leftrightarrow y=2\)
Vậy có 2 cặp x;y thỏa mãn là .........................
25 tháng 12 2016
\(!y-2!\le1\Rightarrow1\le y\le3\Rightarrow co.the=\left\{1,2,3\right\}\)
\(!2x-3!\le1\Rightarrow1\le x\le2=>x.cothe.=\left\{1,2\right\}\)
Với x=1,2=>có y=2
với 1,3 không có x thỏa mãn
KL:
(xy)=(1,2); (2,2)
HT
1
14 tháng 2 2016
Ta có: -210=(-1)+(-2)+(-3)+...+(-x+1)+(-x)
-210=-[1+2+3+...+(x-1)+x)]
-210=\(-\left[\frac{x\left(x+1\right)}{2}\right]\)
hay 210=\(\left[\frac{x\left(x+1\right)}{2}\right]\)
=> x(x+1)=210.2=420
=> 20.21=420 => x=20
NS
2
L
2
Số nguyên x chính là : -20