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4 tháng 2 2017

=1(2+1)(2^2+1)...(2^641)-2^128

=(2-1)(2+1)(2^2+1)(2^4+1)...(2^64+1)-2^128

=(2^2-1)(2^2+1)(2^4+1)...(2^64+1)-2^128.......

=(2^64-1)(2^64+1)-2^128

=2^128-1-2^128

=-1

4 tháng 2 2017

\(=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)-2^{128}\)

\(...\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)-2^{128}\)

\(=2^{128}-1-2^{128}=-1\)

14 tháng 7 2016

a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1

b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2

đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)

\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=.......\)

2B=(5^64-3^64)(5^64+3^64)

2B=5^128-3^128

B=(5^128-3^128)/2 (thế vào đề bài)

=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)

14 tháng 7 2016

a) A = ( 2-1)(2+1)(22+1)...(232+1)-264

         =(22-1)(22+1)(24+1)... -264

       =....

       =264-1-264=1

câu b tương tự nhá

5 tháng 11 2017

a, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}=2^{64}-1-2^{64}=-1\)

b,\(B=\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+\dfrac{5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)}{2}+\dfrac{5^{128}-3^{128}}{2}\)\(=\dfrac{\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}\)

\(=\dfrac{\left(5^{64}-3^{64}\right)\left(5^{64}+3^{64}\right)+5^{128}-3^{128}}{2}=\dfrac{2.5^{128}}{2}=5^{128}\)

4 tháng 9 2018

Câu a : Ta có :

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

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

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

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

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

Vậy \(A>B\)

Câu b : Ta có :

\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)}{2}\)

\(=\dfrac{...\left(3^{64}-1\right)\left(3^{64}+1\right)}{2}\)

\(=\dfrac{3^{128}-1}{2}< 3^{128}-1\)

Vậy \(A< B\)

Bài 2:

a: \(\left(a-b-2\right)^2-\left(2a-2b\right)\left(a-b-2\right)+a^2-2ab+b^2\)

\(=\left(a-b\right)^2-4\left(a-b\right)+4+\left(a-b\right)^2-2\left(a-b\right)\left(a-b-2\right)\)

\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left[\left(a-b\right)^2-2\left(a-b\right)\right]\)

\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left(a-b\right)^2+4\left(a-b\right)\)

\(=4\)

b: \(\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)

\(=\left(2^{64}-1\right)\left(2^{64}+1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)

\(=\left(2^{128}-1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)

\(=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)

\(=2^{512}-1+1=2^{512}\)

c: \(24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)

\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}\)

=-1

28 tháng 10 2017

Giúp vs @@Phạm Hoàng GiangTrần Quốc LộcTrần Thị Hươnghattori heijiTRẦN MINH HOÀNGAn Nguyễn BáRibi Nkok NgokKien Nguyen

Trần Đăng NhấtHung nguyen

28 tháng 10 2017

Sửa đề bài 1 : Rút gọn

a,\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right).........\left(2^{32}+1\right)-2^{64}\)