tính nhanh
A= 1/3+1/3^2+1/3^3+...+1/3^8
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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
\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\\ A=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}\)
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
\(a:=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}\right)\\ =1+\dfrac{3}{5}=1\dfrac{3}{5}\\ b:=\left(\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{5}{14}+\dfrac{3}{14}\right)\\ =\dfrac{3}{7}+\dfrac{8}{14}\\ =\dfrac{3}{7}+\dfrac{4}{7}=1\)
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 .`
`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
=>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=\(\frac{3^8-1}{3^8}:2=\frac{3^8-1}{2.3^8}\)