10/5.7+10/7.9+....+10/61.63
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NT
1
14 tháng 3 2016
10/5.7+10/7.9+...+10/61.63
=2.(5/5.7+5/7.9+.....+5/61.63)
5/5.7=5/5-5/7
=5.(5/5.7+5/7.9+....+5/61.63)
=5.(5/5-5/7+5/7-5/9+...+5/61-5/63)
=5.(5/5-5/63)
=5.(315/315-25/315)
=5.290/315
=1450/315=290/63
15 tháng 6 2023
\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
23 tháng 7 2021
Ta có: \(1-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-...-\dfrac{2}{61\cdot63}-\dfrac{2}{63\cdot65}\)
\(=1-\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{61\cdot63}+\dfrac{2}{63\cdot65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{61}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
NH
1
14 tháng 5 2017
\(M=\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{61.63}\)
\(2M=2.\left(\frac{3}{5.7}+\frac{3}{7.9}+.....+\frac{3}{61.63}\right)\)
\(2M=3.\left(\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{61.63}\right)\)
\(2M=3.\left(\frac{1}{5}-\frac{1}{63}\right)\)
\(2M=\frac{3.58}{315}=\frac{58}{105}\)
\(M=\frac{58}{105}.\frac{1}{2}=\frac{29}{105}\)
28 tháng 5 2017
Ta có thể vt gọn thành :
M = \(\frac{3}{2}\).( \(\frac{1}{5}\)\(-\)\(\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\)... \(+\frac{1}{61}-\frac{1}{63}\))
M = \(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{63}\right)\)
M = \(\frac{3}{2}.\frac{58}{315}\)
M = \(\frac{29}{105}\)
1 tháng 7 2017
Ta có: \(B=1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{61.63}+\dfrac{2}{63.65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}=\dfrac{133}{195}.\)
Vậy \(B=\dfrac{133}{195}.\)
AL
0