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A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\)\(\frac{1}{2^{2019}}\)
2A= \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2019}}\right)\)
2A= \(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\)
2A-A=\(\left(1+2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2019}}\right)\)
A= \(2-\frac{1}{2^{2019}}\)
A=\(\frac{2^{2020}}{2^{2019}}-\frac{1}{2^{2019}}\)
A=\(\frac{2^{2020}-1}{2^{2019}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A-A=2-\frac{1}{2^{2012}}\Rightarrow A=2-\frac{1}{2^{2012}}\)
\(A=\frac{2^{2013}}{2^{2012}}-\frac{1}{2^{2012}}=\frac{2^{2012}+1}{2^{2012}}\)
A= 1+ 1/2 + 1/22 + ... + 1/22012
(1/2)A= 1/2+1/22+...+1/22013
A-(1/2)A= (1+ 1/2 + 1/22 + ... + 1/22012) - ( 1/2+1/22+...+1/22013)
(1/2)A = 1 - 1/22013
A= (1- 1/22013) : 1/2
A= 2 - 1/22012
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(\Leftrightarrow\)\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)
\(\Rightarrow\)\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2012}}\right)\)
\(\Rightarrow\)\(A=2-\frac{1}{2^{2012}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
=>2A=2+1+1/2+1/22+...+1/22011
=>2A-A=(2+1+1/2+1/22+...+1/22011)-(1+1/2+1/22+1/23+...+1/22012)
=>A=2-1/22012
Bài 2 : Rút gọn biểu thức
A = 1 + 1/2 + 1/22 + 1/23 + ... + 1/22012
=>2A=2+1+1/2+1/22+...+1/22011
=>2A-A=(2+1+1/2+1/22+...+1/22011)-(1+1/2+1/22+1/23+...+1/22012)
=>A=2-1/22012
a3+b3+3a2b+3ab2 = (a+b)3 là hằng đẳng thức đáng nhớ "Lập phương của 1 tổng" ở lớp 8.
Có A=20122013+2/20122013-1
=(20122013-1)+3/20122013-1
=20122013-1/20122013-1 + 3/20122013-1
=1 + 3/20122013-1
Có B=20122013/20122013-3
=(20122013-3)+3/20122013-3
=20122013-3/20122013-3 + 3/20122013-3
=1 + 3/20122013-3
Vì 1 + 3/20122013-1>1+20122013-3
nên A>B
Vậy A>B
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\\ 2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}\\ 2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\right)\\ A=2-\dfrac{1}{2^{2012}}\)