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7 tháng 10 2017

\(100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)

\(=1.199+1.195+...+1.3\)

\(=199+195+....+3\)

\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)

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

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

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

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

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

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

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

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

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

\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)

20 tháng 10 2021

các bn giúp mình nhé

20 tháng 10 2021

chụp khó nhìn quá bn ơi

a: A=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)

=100+99+98+...+2+1

=5050

b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

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

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1

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

a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+...+2+1\)

=5050

b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

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

\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)

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

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

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

20 tháng 2 2022

a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=199+195+...+3\)

\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)

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

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

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

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

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

\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)

\(=2c^2\)

18 tháng 9 2021

\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)

\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)

22 tháng 10 2020

a) \(=\left(127+73\right)^2=200^2=40000\)

b) \(=18^8-\left(18^8-1\right)=1\)

c) \(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+...+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+...+2+1=5050\)

d) biến đổi thành \(20^2-19^2+18^2-17^2+..+2^2-1^2\)

rồi giải ra như trên

22 tháng 10 2020

a) Ta có : 2005.2007 = (2006 - 1)(2006 + 1) = 20062 - 12 = 20062 - 1 ( cái khúc này sửa : 2005.2001 thành 2005.2007)

Mà B = 20062

=> 20062 - 1 < 20062 

=> A < B

b) Ta có : B = (2 + 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)

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

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

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

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

Mà C = 232

=> B < C 

c) Tương tự như câu b