Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)
Lời giải:
\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)
\(=5\times \left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{200-199}{199\times 200}\right)\)
\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)
\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
`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`
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=\dfrac{3}{4}\)
\(\Leftrightarrow2\left(x-\dfrac{1}{3}\right)\left(1-\dfrac{1}{10}\right)=\dfrac{3}{4}\Leftrightarrow\dfrac{9}{10}\left(x-\dfrac{1}{3}\right)=\dfrac{3}{8}\)
\(\Leftrightarrow x-\dfrac{1}{3}=\dfrac{5}{12}\Leftrightarrow x=\dfrac{5}{12}+\dfrac{1}{3}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(H=0,25\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{19\times20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{20}\right)=0,25\times\dfrac{19}{20}=\dfrac{19}{80}\)
\(H=\dfrac{0.25}{1\cdot2}+\dfrac{0.25}{2\cdot3}+...+\dfrac{0.25}{199\cdot200}\)
\(=\dfrac{1}{4}\cdot\dfrac{199}{200}=\dfrac{199}{800}\)
\(C=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{2019\times2020}\)
\(=2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{2019\times2020}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\)
\(=2\left(1-\dfrac{1}{2020}\right)=2.\dfrac{2019}{2020}=\dfrac{2019}{1010}\)
\(\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\right)\times x=\dfrac{3}{4}\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\times x=\dfrac{3}{4}\)
\(\left(1-\dfrac{1}{10}\right)\times x=\dfrac{3}{4}\)
\(\dfrac{9}{10}\times x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}\times\dfrac{10}{9}\)
\(x=\dfrac{5}{6}\)
\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)