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= 2 x [1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 +1/7 -1/9 + .., +1/99 - 1/101
= 2 x [ 1 - 1/101 ]
= 2 x 100/101
= 200/101
t cho mik nha
\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+.........+\(\frac{2}{99.101}\)
=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\)
= 1 - \(\frac{1}{101}\)= \(\frac{100}{101}\)
2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5
2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5
2\3x-780\11:[13\2.(1\3-1\11)]=-5
2\3x-780\11:[13\2.8\33]=-5
2\3x-780\11:52\33=-5
2\3x-525\13=-5
2\3x=-5+525\13
2\3x=460\13
x=460\13:2\3
x=690\13
Gọi 2/3.5 +2/5.7 +2/7.9 +...+2/97.99 là A
A=2/3.5 +2/5.7 +2/7.9+...+ 2/97.99
A= 1.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
A=1.(1/3-1/99)
A=1.32/99
A=32/99
Ta có: A>8/25
=>32/99>8.25
Vậy 2/3.5+2/5.7+2/7.9+...+2/97.99>8/25
k cho mk nha!!!
A =(1/2 +1)×(1/3 +1)×(1/4 +1)×....×(1/99 +1)
=3/2x4/3x...............x100/99
=2-1/99
=197/99
A= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{100}{99}\)
A=\(\frac{\left(3\cdot4\cdot5\cdot....\cdot99\right)\cdot100}{2\cdot\left(3\cdot4\cdot5\cdot...\cdot99\right)}\)
A=\(\frac{100}{2}=50\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=> \(\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)>\(\frac{32}{100}\)=32%
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
A= 1/3-1/5+1/5-1/7+....+1/25-1/27
=1/3-1/27
=8/27
vậy A=8/27
\(Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\)\(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
\(Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(Q=\frac{1}{3}+0+0+0-\frac{1}{11}\)
\(Q=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
Ta có :
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{67.69}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{67}-\frac{1}{69}\)
\(=\frac{1}{3}-\frac{1}{69}\)
\(=\frac{22}{69}\)
Mà \(\frac{22}{69}>\frac{11}{35}\left(=\frac{22}{70}\right)\)
\(\Rightarrowđpcm\)
Ta có: \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{67.69}\)
= \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{67}-\frac{1}{69}\)
= \(\frac{1}{3}-\frac{1}{69}\)
= \(\frac{22}{69}\)> 11/35