tìm \(x\) biết: (\(\frac{9}{2.3}+\frac{9}{3.4}+\cdots+\frac{9}{199.200}\) )\(\) .\(x\) =\(\frac25\)
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TD
1
24 tháng 4 2023
(92.3+93.4+...+9199.200).x=25(92.3+93.4+...+9199.200).�=25
⇒[9.(12−13+13−14+...+1199−1200)]x=25⇒[9.(12-13+13-14+...+1199-1200)]�=25
⇒[9.(12−1200)]x=
S
20 tháng 2 2019
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{199\cdot200}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(A=1-\frac{1}{200}\)
\(A=\frac{199}{200}\)
12 tháng 5 2015
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}+\frac{1}{2}x=\frac{3}{2}\)
\(\Leftrightarrow1-\frac{1}{200}+\frac{1}{2}x=\frac{3}{2}\)
\(\Leftrightarrow\frac{199}{200}+\frac{1}{2}x=\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{3}{2}-\frac{199}{200}\)
\(\Leftrightarrow\frac{1}{2}x=\frac{101}{200}\)
\(\Leftrightarrow x=\frac{101}{200}:\frac{1}{2}\)
\(\Leftrightarrow x=\frac{101}{100}\)
12 tháng 5 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}=1-\frac{1}{200}=\frac{199}{200}\)
=> \(\frac{199}{200}+\frac{1}{2}x=1\frac{1}{2}=\frac{3}{2}\Rightarrow\frac{1}{2}x=\frac{101}{200}\Rightarrow x=\frac{101}{100}\)
đúng nhé
16 tháng 4 2018
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(A=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9.\left(1+\left[-\frac{1}{2}+\frac{1}{2}\right]+\left[-\frac{1}{3}+\frac{1}{3}\right]+...+\left[-\frac{1}{99}+\frac{1}{99}\right]-\frac{1}{100}\right)\)
\(A=9.\left(1+0+0+...+0-\frac{1}{100}\right)\)
\(A=9.\left(1-\frac{1}{100}\right)\)
\(A=9.\left(\frac{100}{100}-\frac{1}{100}\right)=9.\left(\frac{99}{100}\right)\)
\(A=\frac{891}{100}=8\frac{91}{100}\)
k cho mk nha
9 tháng 5 2017
\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=\frac{9.1}{1.2.1}+\frac{9.1}{2.3.1}+...+\frac{9.1}{98.99.1}+\frac{9.1}{99.100.1}\)
\(A=1\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(A=1.\frac{99}{100}\)
\(A=\frac{99}{100}\)
8 tháng 8 2016
A=9.(1/1.2+1/2.3+1/3.4+....+1/98.99+1/99.100)
A=9.(1/1-1/2+1/2-1/3+...+1/98-1/99+1/99-1/100)
A=9.(1-1/100)
A=9.99/100
A=901/100
14 tháng 2 2016
=9.(1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100)
=9.(1/1-1/2+1/2-1/3+1/3-1/4+....+1/98-1/99+1/99-1/100)
=9.(1/1-1/100)
=9-9/100
=891/100
TN
23 tháng 4 2016
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9\left(1-\frac{1}{100}\right)\)
\(=9\times\frac{99}{100}\)
\(=\frac{891}{100}\)
23 tháng 4 2016
A=9.(1/1.2 +1/2.3 +1/3.4+...+1/98.99 +1/99.100
A=9.(1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)
A=9.(1-1/100)
A=9.99/100
A=891/100
25 tháng 3 2015
Ta có:
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...\frac{9}{98.99}+\frac{9}{99.100}\)
\(=9.\frac{1}{1.2}+9.\frac{1}{2.3}+9.\frac{1}{3.4}+...+9.\frac{1}{98.99}+9.\frac{1}{99.100}\)
\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=9.\left(1-\frac{1}{100}\right)\)
\(=9.\frac{99}{100}\)
\(=\frac{9.99}{100}\)
\(=\frac{891}{100}\)
`(9/2.3 + 9/3.4 + ...+ 9/ 199.200).x= 2/5`
`9.(1/2.3 + 1/3.4 + ....+ 1/199.200).x=2/5`
`9.(1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/199 - 1/200).x=2/5`
`9.(1/2 - 1/200).x=2/5`
`9.(100/200 - 1/200).x=2/5`
`9. 99/200.x=2/5`
`891/200.x=2/5`
`x=2/5:891/200`
`x=2/5 . 200/891`
`x=80/891`
Ta có: \(x\left(\frac{9}{2\cdot3}+\frac{9}{3\cdot4}+\cdots+\frac{9}{199\cdot200}\right)=\frac25\)
=>\(9x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{199\cdot200}\right)=\frac25\)
=>\(9x\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{199}-\frac{1}{200}\right)=\frac25\)
=>\(9x\left(\frac12-\frac{1}{200}\right)=\frac25\)
=>\(9x\cdot\frac{99}{200}=\frac25\)
=>\(x\cdot\frac{891}{200}=\frac25\)
=>\(x=\frac25:\frac{891}{200}=\frac25\cdot\frac{200}{891}=\frac{2\cdot40}{891}=\frac{80}{891}\)