tính giá trị của biểu thức
G=(1-1/4)(1-1/9)(1-1/16)...(1-1/2025)
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HT
1
XO
28 tháng 7 2020
M = \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}.\right)\left(1-\frac{1}{16}\right)....\left(1-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}=\frac{3.8.15...9999}{4.9.16....10000}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)....\left(99.101\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)....\left(100.100\right)}\)
\(=\frac{\left(1.2.3...99\right).\left(3.4.5..101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)
MK
5
5 tháng 10 2017
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1x64}{2x64}+\frac{1x32}{4x32}+\frac{1x16}{8x16}+\frac{1x8}{16x8}+\frac{1x4}{32x4}+\frac{1x2}{64x2}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\)
\(=\left(\frac{64}{128}+\frac{1}{128}\right)+\left(\frac{32}{128}+\frac{8}{128}\right)+\left(\frac{16}{128}+\frac{4}{128}\right)\)
\(=\frac{65}{128}+\frac{40}{128}+\frac{20}{128}\)
\(=125\)
LT
7
TM
3 tháng 11 2017
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)\(-\frac{1}{64}\)
\(=1-\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}\)\(-\frac{2}{64}-\frac{1}{64}\)
\(=1-\left(\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}-\frac{2}{64}-\frac{1}{64}\right)\)
\(=1-\frac{1}{64}\)
\(=\frac{64}{64}-\frac{1}{64}\)
\(=\frac{63}{64}\)
LN
1
S
10 tháng 7 2018
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{16}\cdot\frac{16.17}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+...+\frac{17}{2}=\frac{1}{2}\left(2+3+...+17\right)=\frac{1}{2}\cdot\frac{16.19}{2}=4.19=76\)
21 tháng 7 2015
C = 8/9 x 15/16 x 24/25 x ... x 99/100
= 2.4/3.3 x 3.5/4.4 x 4.6/5.5 x ... x 9.11/10.10
= (2.3.4...9) x (4.5.6...11) / (3.4.5...10) x (3.4.5...10)
Gian uoc ta co
= 2.11/10.4
= 11/20
NC
0
BN
1
NA
Nguyễn An Ninh
VIP
9 tháng 5 2023
Q = 1+ \(\dfrac{1}{2}\) .(1+2)+ \(\dfrac{1}{3}\) . (1 + 2 + 3) +...+ \(\dfrac{1}{16}\) (1+2+3+...+16)
Q = = 1 + \(\dfrac{1}{2}\) .\(\dfrac{2.3}{2}\) + \(\dfrac{1}{3}\) .\(\dfrac{3.4}{2}\) +...+ \(\dfrac{1}{16}\). \(\dfrac{16.17}{2}\)
Q = \(\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
Q = \(\dfrac{1}{2}\left(3+4+...+17\right)\)
Q = 76
