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= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ ..... + \(\frac{1}{97}\)- \(\frac{1}{99}\)
= 2 . ( \(\frac{1}{3}\)- \(\frac{1}{99}\))
= 2 . \(\frac{2}{3}\)
= \(\frac{4}{3}\)
32% = \(\frac{32}{100}\)= \(\frac{8}{25}\)
\(\frac{4}{3}\)> \(\frac{8}{25}\)=> \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+ \(\frac{2}{7.9}\)+ ..... + \(\frac{2}{97.99}\)> 32%
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(A=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}=\frac{800}{2475}\)
\(32\%=\frac{8}{25}=\frac{792}{2475}\)
\(\frac{800}{2475}>\frac{792}{2475}\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}>32\%\)
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!!!
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
Tự tính
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(M=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{97}-\frac{1}{99}\right).\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)=\frac{1}{2}x\frac{32}{99}=\frac{32}{198}\)
bn tự rút gọn nha mk mới làm tắt đó
F = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
F = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
F = \(\frac{1}{3}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{9}-\frac{1}{9}\right)-...-\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
F = \(\frac{1}{3}-\frac{1}{99}\)
F = \(\frac{32}{99}\)
\(F=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(\Rightarrow F=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow F=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow F=\frac{32}{99}\)
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%
Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{97.99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+......+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
\(M=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
\(M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(M=\frac{1}{3}-\frac{1}{99}\)
\(M=\frac{32}{99}\)
M=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
M=\(\frac{1}{3}-\frac{1}{99}\)
M=\(\frac{32}{99}\)
TICK ỦNG HỘ NHA
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Đặt : \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Do \(\frac{32}{99}>32\%\)nên \(A>32\%\left(đpcm\right)\)
7/15=1/5+4/15