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\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)
\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)
⇒\(A=2-\)\(\dfrac{1}{2^{2012}}\)
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\(A=\frac{1}{30}+\frac{1}{42}+....+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
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A=2^2005-(2^2004+2^2003+...+2+1)
đặt A=2^2004+2^2003+...+2+1
2A=2(1+2+...+2^2004)
2A=2+2^2+2^2005
2A=2+2^2+...+2^2005
A=2^20052005-1 khi đó A=2^2005-A
suy ra A=2^2005-(2^2005-1)=2^2005-2^2005+1=1
CHÚC BẠN HỌC TỐT ##
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=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023
=0+0+...+0-1-2023
=-2024
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\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)\
\(A=\frac{1}{2014}\)
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(=\frac{2}{\left(2.3\right).2}+\frac{2}{\left(6.5\right).2}+\frac{2}{\left(10.7\right).2}+...+\frac{2}{\left(198.101\right).2}\)
\(=\frac{2}{2.\left(3.2\right)}+\frac{2}{6.\left(5.2\right)}+\frac{2}{10.\left(7.2\right)}+...+\frac{2}{198.\left(101.2\right)}\)
\(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+...+\frac{2}{198.202}\)
\(=\frac{4}{2.6}:2+\frac{4}{6.10}:2+\frac{4}{10.14}:2+...+\frac{4}{198.202}:2\)
\(=\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{198.202}\right):2\)
\(=\left(\frac{1}{2}-\frac{1}{202}\right):2\)
\(=\frac{50}{202}=\frac{25}{101}\)
Vậy \(A=\frac{25}{101}\)
frac,left,right là gì vậy ?