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cách này ngon hơn nè
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
Ta có : \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{1}{2012}\)(sửa lại đề)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=1+\left(\frac{2011}{2}+1\right)+...+\left(\frac{1}{2012}+1\right)\)(2012 số hạng 1)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=1+\frac{2013}{2}+...+\frac{2013}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
=> x = 2013
Vậy x = 2013
(1+2012/1)(1+2012/2)(1+2012/3).......(1+2012/1000)
(1+1000/1)(1+1000/2)..........(1+1000/2012)
Tính A
A= 1 + 1/2 + 1/2^3 + 1/2^3 +... + 1/2^2012
1/2.A=1/2 + 1/2^3 + 1/2^3 +... + 1/2^2012+1/2^2013
A-1/2.A=(1 + 1/2 + 1/2^3 + 1/2^3 +... + 1/2^2012)-(1/2 + 1/2^3 + 1/2^3 +... + 1/2^2012+1/2^2013)
1/2.A=1-1/2^2013
A=2-1/2^2012
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
tính chỗ có gạch trước nhá
chỗ +...+ là ko biết gì nên tớ cho thành cộng luôn nha
( 2 x 2012 ) : [ 1 : ( 1 + 2 )] + [ 1 : ( 1 + 2 + 3 )] + [ 1 : ( 1 + 2 +....+ 2011 + 2012 )] =
Số các số hạng của dãy 1+..+2012 là : ( 2012 - 1 ) : 1 + 1 = 2012 | 1 + ... + 2012 = ( 2012 + 1 ) x 2012 : 2 = 2025078
4024 : [ 1 : 3 ] + [ 1 : 6 ] + [ 1 : 2025078 ] =
4024 x 3 + 1 + 1 = 1 1 6 2025078
12072 + 1 + 1 = 120721 + 1 = 12072 337514 6 2025078 6 2025078 2025078
==========>NHỚ K MÌNH , CHÚC HỌC GIỎI<===========
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}\)
\(\Rightarrow2A-A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2013}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)
\(\Rightarrow A=-1+\frac{1}{2^{2013}}=\frac{-2^{2013}+1}{2^{2013}}\)