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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
Vì: \(2^4\)có tận cùng là đặc biệt
Ta có: \(2^{2013}=2^{4.503+1}=\left(2^4\right)^{503}.2=\overline{....6}^{503}.2=\overline{....2}\)
\(\dfrac{5.4^2+16}{2^3}=\dfrac{16\left(5+1\right)}{2^3}=2.6=12\)
\(\dfrac{5^{16}}{5^{14}}+2^2.2^3=5^2+2^5=25+32=57\)
\(\dfrac{7^{2012}}{7^{2010}}-6^2=7^2-6^2=49-36=13\)
\(2^2.3+\dfrac{250}{5^2}=12+10=22\)
\(2.9.50-2012^0=9.100-1=899\)
\(\dfrac{123}{3}-\dfrac{4^3}{2^4}=41-\dfrac{4^2.4}{2^4}41-4=37\)
A = 2011^2012 - 2011^ 2011 = 2011^2011 . ( 2011 - 1 ) = 2011^2011 . 2010
B = 2011^2013 - 2011^2012 = 2011^2012 . ( 2011 - 1 ) = 2011^2012 . 2010
Vì 2011^2011 < 2012^2011
=> 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 :
\(\hept{\begin{cases}\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\\\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\\\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\end{cases}}\)
\(\Rightarrow P>Q\)