Kết quả của tổng M biết: M= \(\dfrac{2}{1x2}\)+\(\dfrac{2}{2x3}\)+......+\(\dfrac{2}{50x51}\)
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C=\(\dfrac{2}{1x2}\)+\(\dfrac{2}{2x3}\)+\(\dfrac{2}{3x4}\)+...+\(\dfrac{2}{2018x2019}\)+\(\dfrac{2}{2019x2020}\)
\(C=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{2019\times2020}\)
\(=2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{2019\times2020}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\)
\(=2\left(1-\dfrac{1}{2020}\right)=2.\dfrac{2019}{2020}=\dfrac{2019}{1010}\)
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lớp 5 đây á
no no
đây ko phải lớp 5 mọi người nhỉ ?
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Tìm x:
(x -\(\dfrac{1}{3}\) ) x (\(\dfrac{2}{1x2}\)+ \(\dfrac{2}{2x3}\)+ \(\dfrac{2}{3x4}\) + … + \(\dfrac{2}{9x10}\)) = \(\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)
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Bài 3:(nâng cao) tính nhanh:
\(\dfrac{2}{1x2}\) + \(\dfrac{2}{2x3}\) + \(\dfrac{2}{3x4}\) + ........ +\(\dfrac{2}{18x19}\) + \(\dfrac{2}{19x20}\)
\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)
\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\cdot\left(1-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)
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Tìm x biết:
\(\dfrac{x}{x+1}=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{31x32}\)
Trả lời nhanh giúp mìn nhé
`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`
`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`
`=>x/(x+1)=1-1/32`
`=>x/(x+1)=31/32`
`=>32x=31(x+1)`
`=>32x=31x+31`
`=>32x-31x=31`
`=>x=31`
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D=\(\dfrac{5}{1x2}\)+\(\dfrac{5}{2x3}\)+\(\dfrac{5}{3x4}\)+....+\(\dfrac{5}{199x200}\)
\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)
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Lời giải:
\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)
\(=5\times \left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{200-199}{199\times 200}\right)\)
\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)
\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)
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H=\(\dfrac{0,25}{1x2}\)+\(\dfrac{0,25}{2x3}\)+......+\(\dfrac{0,25}{199x200}\)
\(H=0,25\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{19\times20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{20}\right)=0,25\times\dfrac{19}{20}=\dfrac{19}{80}\)
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\(H=\dfrac{0.25}{1\cdot2}+\dfrac{0.25}{2\cdot3}+...+\dfrac{0.25}{199\cdot200}\)
\(=\dfrac{1}{4}\cdot\dfrac{199}{200}=\dfrac{199}{800}\)
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\(\dfrac{1}{2x3}\)+\(\dfrac{1}{1x2}\)+..................+\(\dfrac{1}{^{\text{64x65}}}\)
B= \(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+\(\dfrac{1}{3x4}\)+.....+\(\dfrac{1}{198x199}\)+\(\dfrac{1}{199x200}\)
\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)
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E=\(\dfrac{0,5}{1x2}\)+\(\dfrac{0,5}{2x3}\)+\(\dfrac{0,5}{3x4}\)+......+\(\dfrac{0,5}{198x199}\)+\(\dfrac{0,5}{199x200}\)
\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
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