Tìm x biết
2.x+2/3.5+2/5.7+2/7.9+2/9.11= -2016/2017
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TT
1
TT
1
HT
27 tháng 8 2015
\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)
\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)
\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)
\(x=\frac{4}{5}+\frac{1}{5}\)
\(x=1\)
NV
1
S
26 tháng 7 2016
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
OY
22 tháng 7 2021
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
ND
10 tháng 5 2022
\(S=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{1}-\dfrac{1}{11}=\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)
GM
1
3 tháng 4 2017
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
VT
11
1 tháng 5 2018
( \(\frac{1}{1x3}\)+ \(\frac{1}{3x5}\)+....+\(\frac{1}{9x11}\)) x \(y\) = \(\frac{2}{3}\)
( \(\frac{2}{1x3}\)+ \(\frac{2}{3x5}\)+...+\(\frac{2}{9x11}\)) x \(y\) = \(\frac{4}{3}\) (nhân 2 vế lên với 2)
(1 - \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)- ...+ \(\frac{1}{9}\)- \(\frac{1}{11}\)) x \(y\)= \(\frac{4}{3}\)
( 1 - \(\frac{1}{11}\)) x \(y\)=\(\frac{4}{3}\)
\(\frac{10}{11}\) x \(y\) =\(\frac{4}{3}\)
\(y\) = \(\frac{4}{3}\): \(\frac{10}{11}\)
\(y\) = \(\frac{4}{3}\)x \(\frac{11}{10}\)
\(y\) =\(\frac{22}{15}\)
28 tháng 4 2021
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
AL
0
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
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