tìm y
[ 1/1.3+1/3.5+1/5.7+1/7.9 ] . y = 2/3
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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 \(\frac{11}{10}\)
\(y\) =\(\frac{22}{15}\)
NV
1
S
26 tháng 7 2016
\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)
\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)
\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)
\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)
\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)
\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)
TT
1
TT
1
HT
27 tháng 8 2015
\(\left[\frac{12}{11}-\left(\frac{1}{2}+\frac{1}{44}\right)\right].\left(x-0,2\right)=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{25}{44}.\left(x-0,2\right)=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{9.11}\right)\)
\(x-0,2=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right):\frac{25}{44}\)
\(x-\frac{1}{5}=\frac{22}{25}.\left(1-\frac{1}{11}\right)=\frac{22}{25}.\frac{10}{11}=\frac{4}{5}\)
\(x=\frac{4}{5}+\frac{1}{5}\)
\(x=1\)
DN
0
TT
0
NP
1
VQ
30 tháng 3 2023
A= \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{97.99}\)
2A= 1 - \(\dfrac{1}{3}\)+\(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+\(\dfrac{1}{97}\)-\(\dfrac{1}{99}\)
2A= 1-\(\dfrac{1}{99}\)
2A= \(\dfrac{98}{99}\)
A= \(\dfrac{98}{99}\) : 2
A=\(\dfrac{49}{99}\)
NP
1
S
30 tháng 3 2023
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(=\dfrac{49}{99}\)
1 tháng 9 2016
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(S=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}=\frac{1}{8}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
\(S=\frac{1}{2}.\left(\frac{58}{45}\right)\)
\(S=\frac{29}{45}\)
giúp mình với
\(\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right].y=\frac{2}{3}\)
\(\Leftrightarrow\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right].y=\frac{2}{3}\)
\(\Leftrightarrow\left[\frac{1}{1}-\frac{1}{9}\right].y=\frac{2}{3}\)
\(\Leftrightarrow\frac{8}{9}.y=\frac{2}{3}\)
\(\Leftrightarrow y=\frac{2}{3}:\frac{8}{9}\)
\(\Leftrightarrow y=\frac{3}{4}\)