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Ta có: \(A=1+3^2+3^4+3^6+...+3^{2008}\)
\(\Leftrightarrow3A=3^1+3^3+3^5+...+3^{2009}\)
\(\Leftrightarrow A+3A=1+3^1+3^2+3^3+...+3^{2009}\)
\(\Leftrightarrow4A=1+3^1+3^2+...+3^{2009}\)
\(\Leftrightarrow12A=3^1+3^2+3^3+...+3^{2010}\)
\(\Leftrightarrow12A-4A=3^{2010}-1\)
\(\Leftrightarrow8A=3^{2010}-1\)
Lại có: B=8A-32010
\(\Leftrightarrow B=3^{2010}-1-3^{2010}=0-1=\left(-1\right)\)
Vậy B=(-1)
A = 1+32+34+..........+32008
=> A = 30+32+34+.......+32008
=> 9A = 32+34+36+.........+32010
=> 9A -A= 32+34+36+.........+32010- 30+32+34+.......+32008
=> 8A = 32010- 1
=> 8A -32010= 32010- 1 -32010
=> 8A -32010 = -1
=> B = -1
\(A=1+3^2+3^4+...+3^{2008}\)
\(9A=3^2+3^4+...+3^{2008}+3^{2010}\)
\(\Rightarrow8A=3^{2010}-1\)
\(\Rightarrow B=3^{2010}-1-3^{2010}=-1\)
\(Q=\frac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Ta có :
\(\hept{\begin{cases}\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\\\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\\\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\end{cases}}\)
\(\Rightarrow P>Q\)
A=1+3+32+33+......+32008
32A=32+33+34+35+......+32010
9A-A=(32+33+34+35+......+32010)-(1+3+32+33+.....+32008)
8A=32010-(1+3)
8A=32010-4
B=8A-32010
-->B=32010-4-32010
B= -4
Vậy B= -4
Bài 1:
1. 36.7 + 34.37 + 19.100
= 34.(32.7 + 37) + 19.100
= 81.100 + 19.100
= 100.(81 + 19)
= 100.100
= 10000
2) 2.14.98+7.4.32-28.30
= 28.98 + 28.32 - 28.30
= 28. (98 + 32 - 30)
= 28.100
= 2800
3) (56.35+56.18):53
= [56.(35 + 18)] : 53
= 56.53:53
= 56
4) (158.129-158.39):28
= [158. (129 - 39)] : 28
= 158.90:28
= 5,675
9A=\(3^2+3^4+3^6+...+3^{2012}\)
9A-A=\(\left(3^2+3^4+3^6+...+3^{2012}\right)-\left(1+3^2+3^4+...+3^{2010}\right)\)
8A=\(\left(3^2+3^4+3^6+...+3^{2010}\right)+3^{2012}-1-\left(3^2+3^4+3^6+...+3^{2010}\right)\)
8A=\(3^{2012}-1\)
=>8A-\(3^{2012}\)=-1