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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
\(Q=\frac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
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+2+22+23+...+22011+22012
2A = 2+22+23+24+.....+22012+22013
2A - A = ( 2+22+23+24+....+22012+22013 ) - ( 1+2+22+23+.....+22011+22012 )
A = 22013 - 1 < 22013
=> A < B
1. Ta có :
\(4A=\frac{2^2\left(2^{18}-3\right)}{2^{20}-3}=\frac{2^{20}-12}{2^{20}-3}=\frac{2^{20}-3-9}{2^{20}-3}=\frac{2^{20}-3}{2^{20}-3}-\frac{9}{2^{20}-3}=1-\frac{9}{2^{20}-3}\)
\(4B=\frac{2^2\left(2^{20}-3\right)}{2^{22}-3}=\frac{2^{22}-12}{2^{22}-3}=\frac{2^{22}-3-9}{2^{22}-3}=\frac{2^{22}-3}{2^{22}-3}-\frac{9}{2^{22}-3}=1-\frac{9}{2^{22}-3}\)
Vì \(2^{20}-3< 2^{22}-3\)
\(\Leftrightarrow\frac{9}{2^{20}-3}>\frac{9}{2^{22}-3}\)
\(\Leftrightarrow1-\frac{9}{2^{20}-3}< 1-\frac{9}{2^{22}-3}\)
\(\Leftrightarrow4A< 4B\)
\(\Leftrightarrow A< B\)
Vậy...
b/ Tương tự
2. TA CÓ: D=\(\frac{2011+2012}{2012+2013}\)
=\(\frac{2011}{2012+2013}+\frac{2012}{2012+2013}\)
VÌ 2012+2013>2012
MÀ \(\frac{2011}{2012+2013}<\frac{2011}{2012}\)(1)
VÌ 2012+2013>2013
MÀ \(\frac{2012}{2012+2013}<\frac{2012}{2013}\)(2)
TỪ (1) VÀ (2) \(\Rightarrow\frac{2011+2012}{2012+2013}<\frac{2011}{2012}+\frac{2012}{2013}\)
VẬY C > D
\(M=1+2+2^2+.....+2^{2012}\)
\(\Rightarrow2M=2+2^2+2^3+....2^{2013}\)
\(\Rightarrow2M-M=M=2^{2013}-1\)
\(\Rightarrow M=N\left(=2^{2013}-1\right)\)
2M=2+22+23+.....+22013
2M-M=22013-1
Suy ra M=N vì M=22013-1 bằng N=22013-1