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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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\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)
\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)
⇒\(A=2-\)\(\dfrac{1}{2^{2012}}\)
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\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)\
\(A=\frac{1}{2014}\)
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a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
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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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a) 32 . 53 + 92 = 9 . 125 + 81
= 1 125 + 81 = 1 206
b) 83 : 42 - 52 = 512 : 16 - 25 = 32 - 25 = 7
c) 33 . 92 - 52.9 + 18 : 6 = 27 . 81 - 25 . 9 + 3
= 2 187 - 225 + 3 = 1 962 + 3 = 1 965
\(A=\frac{1,6:\left(1+2\right)}{3}\)
\(A=\frac{1,6:3}{3}\)
\(A=\frac{8}{45}\)
1,6 : 3= 8/15 k cái ủng hộ