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8 tháng 10 2015

(2+1)(22+1)(24+1)(28+1)(216+1)=3(22+1)(24+1)(28+1)(216+1)=(22-1)(22+1)(24+1)(28+1)(216+1)=(24-1)(24+1)(28+1)(216+1)=(28-1)(28+1)(216+1)=(216-1)(216+1)=(232-1)

8 tháng 10 2015

(2+1)(22+1)(24+1)(28+1)(216+1)

=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)

=(22-1)(22+1)(24+1)(28+1)(216+1)

=(24-1)(24+1)(28+1)(216+1)

=(28-1)(28+1)(216+1)

=(216-1)(216+1)

=232-1

7 tháng 8 2017

(22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1) = (24 - 1)(24 + 1)(28 + 1)(216 + 1) = (28 - 1)(28 + 1)(216 + 1) = (216 - 1)(216 + 1) = 232 - 1

26 tháng 3 2018

Ta có: (2+1)(22+1)(24+1)(28+1)(216+1)(232+1)

= (2-1)(2+1)(22+1)(24+1)(28+1)(216+1)(232+1)

= (22-1)(22+1)(24+1)(28+1)(216+1)(232+1) 

= (24-1)(24+1)(28+1)(216+1)(232+1)

= (28-1)(28+1)(216+1)(232+1)

= (216-1)(216+1)(232+1) 

= (232-1)(232-1) 

= 264-1 

26 tháng 3 2018

đặt biểu thức là A. Xong rùi đó, A đơn giản nhất còn gì

17 tháng 9 2018

A = 12 – 22 + 32 – 42 + … – 20042 + 20052

     A = 1 + (32 – 22) + (52 – 42)+ …+ ( 20052 – 20042)

     A = 1 + (3 + 2)(3 – 2) + (5 + 4 )(5 – 4) + … + (2005 + 2004)(2005 – 2004)

     A = 1 + 2 + 3 + 4 + 5 + … + 2004 + 2005

     A = ( 1 + 2002 ). 2005 : 2 = 2011015

b/  B = (2 + 1)(22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = (22  - 1) (22 +1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = ( 24 – 1)(24 + 1)(28 + 1)(216 + 1)(232 + 1) – 264

     B = …

     B =(232 - 1)(232 + 1) – 264

     B = 264 – 1 – 264

     B = - 1

17 tháng 9 2018

xin lỗi nha chỗ câu a mình lộn

chỗ (1+2002)x2005:2=2011015 là sai nha 

       (1+2005)x2005:2= 2011015 là đúng nha 

26 tháng 6 2018

Giải:

a) \(M=\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(\Leftrightarrow3M=2^{32}-1\)

\(\Leftrightarrow M=\dfrac{2^{32}-1}{3}\)

Vậy ...

b) \(N=16\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=48\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^2-1\right)\left(7^2+1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^4-1\right)\left(7^4+1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^8-1\right)\left(7^8+1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=\left(7^{16}-1\right)\left(7^{16}+1\right)\)

\(\Leftrightarrow3N=7^{32}-1\)

\(\Leftrightarrow N=\dfrac{7^{32}-1}{3}\)

Vậy ...

30 tháng 7 2018

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

11 tháng 7 2018

Ta có :

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\)

\(=2^{16}-1\left(đpcm\right)\)

11 tháng 7 2018

\(3\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)

\(=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)\)

\(=\left(2^8-1\right).\left(2^8+1\right)\)

\(=2^{16}-1\)

Vậy \(3.\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)=2^{16}-1\left(đpcm\right)\)

23 tháng 6 2017

Giúp mình làm bài này nhé!!!eoeoeoeoeoeo

23 tháng 6 2017

Ta có:

\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1\)

Vậy...