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

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

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

[(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

\(\dfrac{1}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)=\dfrac{2}{1342671}\)

=1/2.(2/2001.2003+2/2003.2005+.....+2/2011.2013)

=1/2(1/2001-1/2003+1/2003-1/2005+....+1/2011-1/2013)

=1/2(1/2001-1/2013)

=2/1342671

\(A=\frac{2012\times2013-1}{2011\times2013+2012}\)

\(A=\frac{2012\times2013-1}{2011\times2013+2011+1}\)

\(A=\frac{2012\times2013-1}{2013\times\left(2011+1\right)+1}\)

\(A=\frac{2012\times2013-1}{2013\times2012+1}\)

\(A=\frac{2012\times2013+1-2}{2012\times2013+1}\)

\(A=1-\frac{2}{2012\times2013+1}\)

Tính tử số 2012x2013-1=2011x2013+2013-1=2011x2013+2012

Do tử=mẫu =2011x2013=2012

Suy ra: A=1

\(A=\frac{2012\times2013-1}{2011\times2013+2012}\)

\(A=\frac{\left(2011+1\right)\times2013-1}{2011\times2013+2012}=\frac{2011\times2013+2013-1}{2011\times2013+2012}=\frac{2011\times2013+2012}{2011\times2013+2012}=1\)

Trả lời

2012.2013-1/2011.2013+2012

=2011.2013+2012-1/2011.2013+2012

=-1/1

=-1.

Học tốt !

\(\frac{2012.2013-1}{2011.2013+2012}\)

\(=\frac{2012.2013-1}{\left(2012-1\right)\left(2012+1\right)+2012}\)

\(=\frac{2012.2013-1}{2012^2-1^2+2012}\)

\(=\frac{2012.2013-1}{2012.2013-1}\)

\(=1\)

Chúc em học tốt !!!

Mình đang cần gấp

- Số được giảm 100 lần là: 20122 - 100 = 20022

- Tổng của 2 số trên là: 20122 + 20022 = 40144

\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2011.2013}\right)\)

\(A=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{4048144}{2011.2013}\)

\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2012.2012}{2011.2013}\)

\(A=\frac{2.3.4...2012}{1.2.3...2011}.\frac{2.3.4...2012}{3.4.5...2013}\)

\(A=2012.\frac{2}{2013}=\frac{4024}{2013}\)

Ta có :

\(A=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right).................\left(1+\dfrac{1}{2011.2013}\right)\)

\(A=\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\left(\dfrac{8}{8}+\dfrac{1}{8}\right)................\left(\dfrac{9999}{9999}+\dfrac{1}{9999}\right)\)

\(A=\dfrac{4}{3}.\dfrac{9}{8}...............\dfrac{10000}{9999}\)

\(A=\dfrac{4.9.................10000}{3.8.............9999}\)

\(A=\dfrac{2.2.3.3................100.100}{1.3.2.4...............99.101}\)

\(A=\dfrac{2.100}{101}=\dfrac{200}{101}\)

~ Chúc bn học tốt ~