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2 tháng 2 2015

hơi vô lí bạn ạ .Tai neu a=0 thi o=1 sao!

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

21 tháng 7 2021

a) Ta có x + y + z = 0

=> x + y = -z

=> (x + y)3 = (-z)3

=> x3 + y3 + 3xy(x + y) = -z3

=> x3 + y3 + z3 = -3xy(x + y) 

=> x3 + y3 + z3 = -3xy(-z)

=> x3 + y3 + z3 = 3xyz (đpcm) 

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

3 tháng 9 2019

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

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

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

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

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

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

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

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

\(=-1\)

3 tháng 9 2019

\(\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(99^2-100^2\right)\) 

\(=\left(1-2\right)\left(2+1\right)+\left(3-4\right)\left(4+3\right)+....+\left(99-100\right)\left(100+99\right)\) 

\(=\left(-1\right)\left(1+2+3+....+100\right)=\frac{\left(-1\right)100.99}{2}=-4950\)

27 tháng 10 2020

Ta có A = (32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = 8(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = (38 - 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = (316 - 1)(316 + 1)(332 + 1)(364 + 1)

=> 8A = (332 - 1)(332 + 1)(364 + 1)

=> 8A = (364 - 1)(364 + 1)

=> 8A = 3128 - 1 (1)

Đặt B = 3126

=> 8B = 3126 . 8 = 3126.(32 - 1) = 3128 - 3126 (2)

Từ (1)(2) => 8A > 8B 

=> A > B 

18 tháng 10 2015

Phân tích 3=4-1=\(2^2-1\)