-1-1/2-1/4-1/8-...-1/1024
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DK
14 tháng 9 2016
Ta có: \(-1=-2+1;-\frac{1}{2}=-1+\frac{1}{2};-\frac{1}{4}=-\frac{1}{2}+\frac{1}{4};...;-\frac{1}{1024}=-\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\left(-2+1\right)+\left(-1+\frac{1}{2}\right)+\left(-\frac{1}{2}+\frac{1}{4}\right)\)\(+...+\left(-\frac{1}{512}+\frac{1}{1024}\right)\)
\(=-2+1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{512}+\frac{1}{1024}\)
\(=-2+\frac{1}{1024}\)
\(=-\frac{2047}{1024}\)
1 tháng 11 2016
\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)\)
\(=\frac{8}{9}-\frac{8}{9}=0\)
NH
8 tháng 4 2023
a)
`9/2-3`
`=9/2-6/2`
`=3/2`
b)
`8/5xx25/12`
`=10/3`
c)
`4:3/7`
`=4xx7/3`
`=28/3`
6
2
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{97}{98}\cdot\dfrac{98}{99}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot98}{2\cdot3\cdot4\cdot...\cdot99}\)
\(=\dfrac{1}{99}\)
NN
16 tháng 6 2023
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}=\dfrac{100}{2}=50\)
13 tháng 12 2015
S1=1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+1997+(-1998)+(-1999)+2000
S1=(1+4-2-3)+(5+8-6-7)+...+(1997+2000-1998-1999)
S1=0+0+...+0
S1=0
câu 2
S2=1+3+4+5+...+99-(2+4+6+...+100)
S2=51.50-(50.51)
S2=0
tich nha
PG
5
21 tháng 3 2017
1/3 - 1/4 = 4/12 - 3/12 = 1/12
2/7 . 14/5 - 1 = 2.14/7.5 - 1 =4/5 - 1 = -1/5
3/4 - 1/25 x 5 = 3/4 - 1.5/25.1 = 3/4 - 1/5 = 11/20
-8 . ( 6. 1/24 ) = -8 . ( 6.1/24.1 ) = -8 . 1/4 = -8 . 1/4.1 = -2
NT
1
26 tháng 8 2019
Ta có:\(\frac{1}{2}>\frac{1}{8};\frac{1}{3}>\frac{1}{8};...;\frac{1}{6}>\frac{1}{8};\frac{1}{7}+\frac{1}{8}+\frac{1}{9}>\frac{3}{8}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}\)
\(=\frac{8}{8}=1\)
Vậy\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1\)
Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2014}\)
\(\Rightarrow-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(-A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow-2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow-2A-\left(-A\right)=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\right)\)
\(-A=2-\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2^{10}}-2\)