Bài 13:
a) \(n-1\inƯ\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
| \(n-1\) | \(-8\) | \(-4\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(4\) | \(8\) |
| \(n\) | \(-7\) | \(-3\) | \(-1\) | \(0\) | \(2\) | \(3\) | \(5\) | \(9\) |
Vậy \(n-1\in\left\{-7;-3;-1;0;2;3;5;9\right\}\)
b) \(n+3\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)
| \(n+3\) | \(-12\) | \(-6\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(1\) | \(2\) | \(3\) | \(4\) | \(6\) | \(12\) |
| \(n\) | \(-15\) | \(-9\) | \(-7\) | \(-6\) | \(-5\) | \(-4\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(3\) | \(9\) |
Vậy \(n+3\in\left\{-15;-9;-7;-6;-5;-4;-2;-1;0;1;3;9\right\}\)
b) \(\left|x\right|\) < 8
\(\left|x\right|\) = 0;1;2;3;4;5;6;7
\(\Rightarrow\)x= -7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7
-7+(-6) + (-5) + (-4) +(-3)+ (-2)+(-1)+0+1+2+3+4+5+6+7
= (-7+7) + (-6+6)+ (-5+5) + (-4+4)+(-3+3) +(-2+2)+ (-1+1)+0
= 0 + 0 +0 + 0 + 0 + 0 +0 + 0 +0 =0
c)10 < \(\left|x\right|\) < 13
\(\left|x\right|\) = 11;12;
\(\Rightarrow\) x = -11;-12;11;12
-11+(-12)+11+12= (-11+11)+ (-12+12)=0+0=0
e)10\(\le\) \(\left|x\right|\) \(\le\) 13
\(\left|x\right|\)=10;11;12;13
x= -10;-11;-12;-13;10;11;12;13
-10+(-11)+(-12)+(-13)+10+11+12+13
=(-10+10)+ (-11+11)+(-12+12) + (-13+13)=0+0+0+0=0
Bài 11:
1-2+3-4+5-6+......+19-20
= (1-2)+(3-4)+ (5-6)+.......+ (19-20) (có tất cả 10 cặp)
= -1+ (-1)+(-1)+.....+(-1)
=-1. 10 = -10
Ta có thể thấy: A\(⋮\) 2 ; 5
A\(⋮̸\)3 ; 4
Ư(A)= \(\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
11. A=(1-2)+(3-4)+(5-6)+(7-8)+....+(19-20)
=(-1)+(-1)+(-1)+(-1)+....+(-1)
=(-1).10
=-10
Các ước của A(tức -10) là {1;-1;2;-2;5;-5;10;-10}
