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\(\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=9900\)
\(\Leftrightarrow99x+\left(\frac{99-1}{1}+1\right)=9900\)
\(\Leftrightarrow99x=9900-99\)
\(\Leftrightarrow x=99\)
k mk nha
(x + 1) +( x + 2) + ... + (x + 99 )= 9900
=>99x +(99-1/1 + 1 )=9900
=>99x=9900-99
=>x=90
ấn chậm quá
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T = \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}=\)
T = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\)
T = \(\frac{1}{1}-\frac{1}{100}=\)
T = \(\frac{99}{100}\)
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A=1+1/2+1+1/6+1+1/12+...+1+1/90=
=9+1/2+1/6+1/12+...+1/90
1/2+1/6+1/12+...+1/90=
1/1x2+1/2x3+2/3x4+...+1/9x10=
\(=\dfrac{2-1}{1x2}+\dfrac{3-2}{2x3}+\dfrac{4-3}{3x4}+...+\dfrac{10-9}{9x10}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}=\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\Rightarrow A=9+\dfrac{9}{10}=9\dfrac{9}{10}\)
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a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20
10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110
10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110
10 x X - 100 = 110
10 x X = 110 + 100
10 x X = 210
X = 210 : 10
X = 21
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{9900}\)
\(=\frac{3}{1\times2}+\frac{3}{2\times3}+\frac{3}{3\times4}+...+\frac{3}{99\times100}\)
\(\Rightarrow A:3=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{100}\times3=\frac{297}{100}\)
Vậy \(A=\frac{297}{100}\).