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A=1-1/(2013*2014)
B=1-1/(2014*2015)
2013*2014<2014*2015
=>1/2013*2014>1/2014*2015
=>-1/2013*2014<-1/2014*2015
=>A<B
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
A = 2016^2015 +1 / 2016^2014+1 < 2016^2015 + 1 + 2015 / 2016^2014 + 1 + 2015
= 2016^2015 + 2016 / 2016^2014 + 2016
= 2016(2016^2014 + 1 ) / 2016(2016^2013 +1)
= 2016^2014 + 1 / 2016^2013 + 1 = B
=> A < B
Ta có:
\(A=\dfrac{9}{a^{2013}}+\dfrac{7}{a^{2014}}\)
\(=\left(\dfrac{8}{a^{2013}}+\dfrac{1}{a^{2013}}\right)+\left(\dfrac{8}{a^{2014}}-\dfrac{1}{a^{2014}}\right)\)
\(=\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)\)
\(B=\dfrac{8}{a^{2014}}+\dfrac{8}{a^{2013}}\)
\(=\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)
Vì \(\dfrac{1}{a^{2013}}>\dfrac{1}{a^{2014}}\Rightarrow\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}>0\)
\(\Rightarrow\left(\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\right)+\left(\dfrac{1}{a^{2013}}-\dfrac{1}{a^{2014}}\right)>\dfrac{8}{a^{2013}}+\dfrac{8}{a^{2014}}\)
Vậy \(A>B\)
Ta có: A=22+23+...+22013
=>2A=2( 22+23+...+22013)
=>2A=23+24+...+22014
=>2A-A=(23+24+...+22014) - (22+23+...+22013)
=>A=22014-22
=>A=22014-4
mà B=22014
=>A<B ( vì 22014-4<22014)
Ta có: A=2^2+2^3+...+2^2012+2^2013
=) 2A= 2^3+....+2^2012+2^2013+2^2014
=) 2A-A=2^2014- 2^2
=) A= 2^2014-4
Mà B= 2^ 2014
=) A< B
Ta có :
A=\(\frac{9}{a^{2013}}+\frac{7}{a^{2014}}\)
=\(\left(\frac{8}{a^{2013}}+\frac{1}{a^{2013}}\right)+\left(\frac{8}{a^{2014}}-\frac{1}{a^{2014}}\right)\)
=\(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)\)
B=\(\frac{8}{a^{2014}}+\frac{8}{a^{2013}}\)
=\(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)
Vì \(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\)nên\(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}>0\)
=> \(\left(\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\right)+\left(\frac{1}{a^{2013}}-\frac{1}{a^{2014}}\right)>\frac{8}{a^{2013}}+\frac{8}{a^{2014}}\)
Vậy \(A>B\)
Chúc em học tốt
#Thiên_Hy
\(A=\frac{9}{a^{2013}}+\frac{7}{a^{2014}}=\frac{8}{a^{2013}}+\frac{1}{a^{2013}}+\frac{7}{a^{2014}}\)
\(B=\frac{8}{a^{2014}}+\frac{8}{a^{2013}}=\frac{7}{a^{2014}}+\frac{1}{a^{2014}}+\frac{8}{a^{2013}}\)
Ta thấy :
\(\frac{8}{a^{2013}}=\frac{8}{a^{2013}}\)
\(\frac{7}{a^{2014}}=\frac{7}{a^{2014}}\)
\(\frac{1}{a^{2013}}>\frac{1}{a^{2014}}\left(a^{2013}< a^{2014}\right)\)
\(\Rightarrow A>B\)