3/2×5 + 3/5×8 + 3/8×11 + 3/11×14 =
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27 tháng 6 2018
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(=\)\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk giúp !!
TT
1
29 tháng 6 2015
= 1/2x5 + 1/5x8 + 1/8x11 + 1/11x14 + 1/14/17
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17
= 1/2 - 1/17
= 15/34.
Vì 15/34 < 1/2
=>đpcm
16 tháng 1 2022
b: \(27D=3^{14}+3^{17}+...+3^{2024}\)
\(\Leftrightarrow26D=3^{2024}-3^{11}\)
hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)
c: \(25E=-5^4-5^6-...-5^{1002}\)
\(\Leftrightarrow24E=-5^{1002}+5^2\)
hay \(E=\dfrac{-5^{1002}+5^2}{24}\)
Z
0
NH
1
16 tháng 9 2023
a: \(=3\cdot\dfrac{6+14-9}{42}\cdot\dfrac{14}{11}\)
\(=3\cdot\dfrac{11}{42}\cdot\dfrac{14}{11}=1\)
b: \(=\dfrac{19}{5}\cdot\dfrac{11}{10}+\dfrac{9}{5}\cdot\dfrac{7}{3}\)
\(=\dfrac{209}{50}+\dfrac{63}{15}=4.18+4.2=8,38\)
S
12 tháng 3 2023
(Dấu . là dấu nhân)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{119.122}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{119}-\dfrac{1}{122}\)
\(=\dfrac{1}{5}-\dfrac{1}{122}\)
\(=\dfrac{122}{610}-\dfrac{5}{610}\)
\(=\dfrac{117}{610}\)
| #Sahara |
12 tháng 3 2023
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+.....+\dfrac{3}{119.122}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+......+\dfrac{1}{119}-\dfrac{1}{122}\)
\(=\dfrac{1}{5}-\dfrac{1}{122}=\dfrac{122-5}{610}=\dfrac{117}{610}\)
14 tháng 4 2023
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
C
5 tháng 3 2020
a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)
Vậy A<\(\frac{1}{2}\).
b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B< 1-\frac{1}{100}< 1\)
Vậy \(B< 1\).
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}=\frac{3}{7}\)