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a) A = 1 + 2 + 3 + 4+... + 50;
Tổng A có 50 số hạng nên A = (1 + 50).50:2 = 1275,
b) B = 2 + 4 + 6 + 8 + ...+100;
Số số hạng của tổng B là: (100 - 2): 2+1 = 50 (số)
Do đó B = (2 +100).50 : 2 = 2550.
c) C = 1 + 3 + 5 + 7 +... + 99;
Số số hạng của tổng C là: (99 - 1): 2 +1 = 50 (số)
Do đó C = (1 + 99). 50 : 2 = 2500.
d = 2 + 5 + 8 + 11 .... 98
= ( 92 - 2 ) : 3 + 1 = 33
= 33 . ( 98 + 2 ) : 2
= 1650
tick cho tớ với
Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
a) \(1+2+3+...+2021=\dfrac{\left(2021+1\right)\left(\dfrac{2021-1}{1}+1\right)}{2}=2043231\)
b) Giống a
c) \(11+13+15+...131=\dfrac{\left(131+11\right)\left(\dfrac{131-11}{2}+1\right)}{2}=4331\)
`a,`
`31/23-(7/32+8/23)`
`=31/23-7/32-8/23`
`=(31/23-8/23)-7/32`
`=1-7/32=25/32`
`b,`
`38/45-(8/45-17/51-3/11)`
`=38/45-8/45+17/51+3/11`
`= (38/45-8/45)+17/51+3/11`
`=2/3+17/51+3/11`
`=1+3/11=14/11`
`c,`
`(1/3+12/67+13/41)-(79/67-28/41)`
`= 1/3+12/67+13/41-79/67+28/41`
`= 1/3+(12/67-79/67)+(13/41+28/41)`
`= 1/3+(-1)+1=1/3+(-1+1)=1/3+0=1/3`
`d,`
`1/5+(-1/6)+1/7+(-1/8)+1/9+1/8+(-1/7)+1/6+(-1/5)`
`= (1/5+ -1/5)+(-1/6+1/6)+(1/7+ -1/7)+(-1/8 +1/8)+1/9`
`= 0+0+0+0+1/9=1/9 .`
Câu 1:
a) \(-\dfrac{3}{7}-\left(\dfrac{2}{3}-\dfrac{3}{7}\right)=\dfrac{-3}{7}-\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{-2}{3}\)
Câu 2:
b) \(\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{4}{5}-\dfrac{1}{3}\cdot\dfrac{6}{5}\right)=\dfrac{2}{15}:\left[\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\right]=\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{-2}{5}\right)=\dfrac{2}{15}:\dfrac{-2}{15}=\dfrac{2}{-2}=-1\)
a: =(1-2-3+4)+(5-6-7+8)+...+(21-22-23+24)
=0+0+...+0
=0
b: =-7111+53+711-153=-6400-100=-6500
c: =-43(-1-296+296)=-43*(-1)=43
A=1+1/2+1/3+1/6+(1/12+1/15+1/20+1/30)+1/35
=71/35+7/30=95/42
Giải:
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1.2.3.4}{2.3.4.5}\)
\(=\dfrac{1}{5}\)
b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Chúc bạn học tốt!
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>3A=\(1+\frac{1}{3}+...+\frac{1}{3^7}\)
=>3A-A=\(\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
=>2A=\(1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=>A=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)