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Đặt biểu thức là A
\(4xA=\frac{4}{2x4x6}+\frac{4}{4x6x8}+\frac{4}{6x8x10}+\frac{4}{8x10x12}+...+\frac{4}{94x96x98}+\frac{4}{96x98x100}\)
\(4xA=\frac{6-2}{2x4x6}+\frac{8-4}{4x6x8}+\frac{10-6}{6x8x10}+\frac{12-8}{8x10x12}+...+\frac{98-94}{94x96x98}+\frac{100-96}{96x98x100}\)
\(4xA=\frac{1}{2x4}-\frac{1}{4x6}+\frac{1}{4x6}-\frac{1}{6x8}+\frac{1}{6x8}-\frac{1}{8x10}+...+\frac{1}{94x96}-\frac{1}{96x98}+\frac{1}{96x98}-\frac{1}{98x100}\)
\(4xA=\frac{1}{2x4}-\frac{1}{98x100}=\frac{49x50-1}{98x100}\Rightarrow A=\frac{49x50-1}{4x98x100}\)
\(\frac{1}{2.4.6}+\frac{1}{4.6.8}+\frac{1}{6.8.10}+..+\frac{1}{50.52.54}\)
\(=\frac{1}{4}.\left(\frac{1}{2.4}-\frac{1}{4.6}+\frac{1}{4.6}-\frac{1}{6.8}+....+\frac{1}{50.52}-\frac{1}{52.54}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2.4}-\frac{1}{52.54}\right)\)
\(=\frac{1}{4}.\frac{175}{1404}=\frac{175}{5616}\)
Sửa lại chút:
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+10}=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2x\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{110}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{11}\right)=2x\frac{9}{22}=\frac{9}{11}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2009}\)
\(=\frac{1}{\frac{2\cdot\left(1+2\right)}{2}}+\frac{1}{\frac{3\cdot\left(3+1\right)}{2}}+\frac{1}{\frac{4\cdot\left(4+1\right)}{2}}+...+\frac{1}{\frac{2009\cdot\left(2009+1\right)}{2}}\)
\(=\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{2009\cdot2010}\)
\(=2\cdot\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2009\cdot2010}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2010}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=1-\frac{1}{1005}\)
\(=\frac{1004}{1005}\)
1/1+2=3=1/1+2+2=6=1/1+2+3+4=10+3+6=19+1/1+2+3+4=29+3+6+10+19+2009=2076nếu mình làm sai thì nhớ chỉ dùm
nhớ kết bạn với mình nhé
2) \(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{99\times101}+\frac{3}{101\times103}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{101\times103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\frac{101}{103}\)
\(=\frac{303}{206}\)
\(\dfrac{1}{2\times4\times6}+\dfrac{1}{4\times6\times8}+...+\dfrac{1}{96\times98\times100}\\ =\dfrac{1}{8}\times\dfrac{1}{1\times2\times3}+\dfrac{1}{8}\times\dfrac{1}{2\times3\times4}+...+\dfrac{1}{8}\times\dfrac{1}{48\times49\times50}\\ =\dfrac{1}{8}\times\left(\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+...+\dfrac{1}{48\times49\times50}\right)\)
Đặt \(A=\dfrac{1}{1\times2\times3}+\dfrac{1}{2\times3\times4}+...+\dfrac{1}{48\times49\times50}\)
\(2A=\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+...+\dfrac{2}{48\times49\times50}\\ 2A=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\\ 2A=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\\ 2A=\dfrac{1}{2}-\dfrac{1}{2450}\\ 2A=\dfrac{612}{1225}\\ A=\dfrac{306}{1225}\)
Thay vào biểu thức ban đầu được:
\(\dfrac{1}{2\times4\times6}+\dfrac{1}{4\times6\times8}+...+\dfrac{1}{96\times98\times100}\\ =\dfrac{1}{8}\times\dfrac{306}{1225}\\ =\dfrac{153}{4900}\)