\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)

\(A=\left(2012-1\right)\left(2012+1\right)-2012^2=2012^2-1-2012^2=-1\)

a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)

b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)

[(2³ - 5) . (-3)+9] . (2²⁰¹² . 2011 - 2012² . 2011+1)

= [ 3 . ( -3 ) + 9] . (2²⁰¹² . 2011 - 2012² . 2011+1)

=  [ (-9) + 9 ] . (2²⁰¹² . 2011 - 2012² . 2011+1)

= 0 . (2²⁰¹² . 2011 - 2012² . 2011+1)

= 0

[ (2³ - 5) . (-3) + 9 ] . ( 2²⁰¹² . 2011 - 2012² . 2011+1 )

= [ 3 . ( -3 ) + 9] . (2²⁰¹² . 2011 - 2012² . 2011+1 )

= 0 . (2²⁰¹² . 2011 - 2012² . 2011+1) = 0

\(A=2011.2013-2012^2\)

Gọi 2012 là a ta có:

\(2011=a-1;2013=a+1\)

\(\Rightarrow A=\left(a+1\right).\left(a-1\right)-a^2\)

\(\Rightarrow A=a^2-a+a-1-a^2\)

\(\Rightarrow A=a^2-1-a^2\)

\(\Rightarrow A=-1\)

\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2011.2013}\)

\(M=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\right)\)

\(M=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(M=2.\left(1-\frac{1}{2013}\right)\)

\(M=2.\frac{2012}{2013}\)

\(M=\frac{4024}{2013}\)

~Học tốt~

M=2.(2/1.3 + 2/3.5 +2/5.7+...+2/2011.2013)

M=2.(1-1/3 +1/3-1/5 +1/5-1/7+... +1/2011-1/2013)

M=2.(1-1/2013)

M=2.2012/2013

M=4024/2013

\(M=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2011.2013}\)

    \(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2011.2013}\right)\)

    \(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

    \(=2\left(1-\frac{1}{2013}\right)\)

    \(=2.\frac{2012}{2013}\)

     \(=\frac{4024}{2013}\)

Study well ! >_<

\(\frac{4}{1.3}\)+\(\frac{4}{3.5}\)+\(\frac{4}{5.7}\)+\(\frac{4}{7.9}\)+...+\(\frac{4}{2011.2013}\)

= 1+\(\frac{1}{3}\)-\(\frac{1}{3}\)+\(\frac{1}{5}\)-\(\frac{1}{5}\)+\(\frac{1}{7}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)+...+\(\frac{1}{2011}\)+\(\frac{1}{2013}\)

=1+       0          +        0        +        0         +...+        0          +         \(\frac{1}{2013}\)

=1+\(\frac{1}{2013}\)

=\(\frac{2014}{2013}\)

k dùm nha

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{2011\cdot2013}\)

\(=2\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{2011\cdot2013}\right)\)

\(=2\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=2\cdot\left(1-\frac{1}{2013}\right)\)

\(=2\cdot\frac{2012}{2013}\)

\(=\frac{4024}{2013}\)

Đặt A ta có : \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{2011.2013}\)

\(2A=4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}...+\frac{1}{2011.2013}\right)\)

\(2A=4\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(2A=4\left(1-\frac{1}{2013}\right)\)

\(2A=4.\frac{2012}{2013}\)

\(2A=\frac{8048}{2013}\)

\(\Rightarrow A=\frac{4024}{2013}\)

a = 2011.2013

a = 2011.(2012+1)

a = 2011.2012 + 2011

b = 2012.2012

b = (2011+1).2012

b = 2011.2012 + 2012

Vì 2011 < 2012

=> 2011.2012 + 2011 < 2011.2012 + 2012

=> a < b