Ta có : \(6.2^{12}+2^{13}=3.2.2^{12}+2^{13}=3.2^{13}+2^{13}=2^{13}\left(3+1\right)=2^{13}.4=2^{13}.2^2=2^{15}=\left(2^3\right)^5=8^5\)\(3^{10}=\left(3^2\right)^5=9^5\)

Vì 8 < 9

=> 8⁵ < 9⁵ ( vì 5 > 0 )

=> 6.2¹² + 2¹³ < 3¹⁰

Vậy 6.2¹² + 2¹³ < 3¹⁰

a, 2^2x+1 = 8³.16⁵

=> 2^2x+1 = (2³)³ . (2⁴)⁵

=> 2^2x+1 = 2⁹ . 2²⁰

^=> 2^2x+1 = 2²⁹

=> 2x+1 = 29 ( vì cơ số 2 > 0 )

=> 2x = 29 - 1

=> 2x = 28

=> x = 28 : 2

=> x = 14

Vậy x = 14

Ta có : \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)

= 3( \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\) ).\(\dfrac{1}{3}\)

= (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\)).\(\dfrac{1}{3}\)

= \(\left(\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{100-97}{97.100}\right).\dfrac{1}{3}\) = \(\left(\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{100}{97.100}-\dfrac{97}{97.100}\right).\dfrac{1}{3}\) = \(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right).\dfrac{1}{3}\) =\(\left(1-\dfrac{1}{100}\right).\dfrac{1}{3}\)

= \(\dfrac{99}{100}.\dfrac{1}{3}\)

= \(\dfrac{99}{100.3}\)

= \(\dfrac{33}{100}\)

Là số 19

Ta có : B = \(\dfrac{2012+2013}{2013+2014}=\dfrac{2012}{2013+2014}+\dfrac{2013}{2013+2014}\) Ta có :

\(\dfrac{2012}{2013}>\dfrac{2012}{2013+2014}\)( vì 2012 > 0; 0<2013<2013+2014 )

\(\dfrac{2013}{2014}>\dfrac{2013}{2013+2014}\)( vì 2013>0; 0<2014<2013+2014 )

=> \(\dfrac{2012}{2013}+\dfrac{2013}{2014}>\dfrac{2012}{2013+2014}+\dfrac{2013}{2013+2014}\) => A > B

Vậy A > B

10.3546.345.11=134570700

Ta có : A =\(\dfrac{54.107-53}{53.107+54}=\dfrac{53.107+107-53}{53.107+54}=\dfrac{53.107+\left(107-53\right)}{53.107+54}=\dfrac{53.107+54}{53.107+54}=1\)Ta có : B = \(\dfrac{135.269-133}{134.269+135}=\dfrac{134.269+269-133}{134.269+135}=\dfrac{134.269+\left(269-133\right)}{134.269+135}=\dfrac{134.269+136}{134.269+135}=\dfrac{134.269+135}{134.269+135}+\dfrac{1}{134.269+135}=1+\dfrac{1}{134.269+135}\) => B > 1 ( vì \(\dfrac{1}{134.269+135}>0\) )

Mà A = 1

=> A < B

Vậy A < B