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1) 167 + ( -252 ) + 52 + ( -67 )
= 167 + 52 + ( -252 ) + ( - 67 )
= 219 + - ( 252 + 67 )
= 219 + - ( 319 )
= -100
2) ( -215 ) + ( - 115 ) + ( - 80 )
= - ( 215 + 115 ) + ( - 80 )
= - 330 + ( - 80 )
= - ( 330 - 80 )
= - 250
3) 118 + 107 - ( 118 - 93 )
= 118 + ( 107 - 93 )
= 118 + 14
= 132
4) 1 + 5 + 9 + ..... + 97 + 101
= ( 101 : 1) - 4 = 96
= 101 x 96
= 9696
5) 38 - 138 + 250 - 350
= - 100 + - 100
= 0
6) - ( - 357 ) + ( - 27 ) + ( - 32 )
= - ( - 357 ) + - ( 27 + 32 )
= - ( - 357 ) + - 59
= - ( 357 + 59 )
= - 416
7) 40 + ( 139 - 172 + 99 ) - ( 139 + 199 - 172 )
= 40 + ( 100 - 100 )
= 40 + 0
= 40
8) ( -1 ) + 2 + ( - 3 ) + 4 + ( - 99 ) + 100
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
a)
\(S=1-2+3-4+...+99-100\)
\(S=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)( Có 50 cặp số -1 )
\(S=\left(-1\right).50\)
\(S=\left(-50\right)\)
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)
\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)
Bài 1:
a: |-x|<5
=>|x|<5
=>-5<x<5
hay \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)
Tổng là 0
c: 2|x-3|=500
=>|x-3|=250
=>x-3=250 hoặc x-3=-250
=>x=253 hoặc x=-247
d: x-17=-8+(-17)
=>x-17=-25
hay x=-8
M = 1 + (-2 + 3 ) + ( -4 + 5 ) + . . . + (-2000 + 2001) + (-2002) + 2003)
M = 1 + 1+ 1+ . . . + 1 + 1
Dãy số này có 2003 số hạng và có 1001 cặp số có tổng bằng nhau
1001.1+1 = 1002 nên M = 1002