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Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
917749738461936926399639748776398646491639394748947630373937366
\(=\frac{6}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-............+\frac{1}{97}-\frac{1}{99}\right).\\ \)
\(=\frac{6}{2}\left(1-\frac{1}{97}\right)\)
tới đây tính máy là ra luôn
Ta có:
\(A=\frac{6}{5x7}+\frac{6}{7x9}+...\frac{6}{97x99}\)
\(=3x\left(\frac{2}{5x7}+\frac{2}{7x9}+...\frac{2}{97x99}\right)\)
\(=3x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3x\left(\frac{1}{5}-\frac{1}{99}\right)\)
\(=3x\left(\frac{99}{495}-\frac{5}{495}\right)\)
\(=3x\frac{94}{495}=\frac{94}{165}\)
Vậy \(A=\frac{94}{165}\)
\(\frac{6}{5}\)x 7 + \(\frac{6}{7}\)x 9 + .... + \(\frac{6}{97}\)x 99
= \(\frac{6}{5}\) - \(\frac{6}{7}\)+\(\frac{6}{7}\)- \(\frac{6}{9}\)+ ..... + \(\frac{6}{97}\)- \(\frac{6}{99}\)
= \(\frac{6}{5}\) - \(\frac{6}{99}\)
= \(\frac{188}{165}\)
nhớ cho đúng đó
=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2
=(1-1/9)chia 2
=8/9 chia 2
=4/9
A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2008}{6033}\)
A = \(\dfrac{1004}{6033}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)
Đến đây bn tự tính nhé.
Đặt A = 1/3.5 + 1/5.7 + 1/7.9 + ..... + 1/99.101
=> 2A = 2/3.5 + 2/5.7 + 2/7.9 + ..... + 2/99.101
=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101
=> 2A = 1/3 - 1/101
=> 2A = 88/303
=> A = 44/303
\(\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+\dfrac{6}{7\cdot9}+.....+\dfrac{6}{33\cdot35}\)
\(=\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{33\cdot35}\right)\cdot3\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{33}-\dfrac{1}{35}\right)\cdot3\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{35}\right)\cdot3\)
\(=\dfrac{32}{3\cdot35}\cdot3\)
\(=\dfrac{32}{35}\)