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19 tháng 7 2017

\(\left(5^2-1\right)A=12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right).\) ... .\(\left(5^{128}+1\right)\)

=\(12\left(5^4-1\right)\)\(\left(5^4+1\right)\)\(\left(5^8+1\right)\). ... .(5128+1)

=\(12\left(5^8-1\right)\left(5^8+1\right)\). ... .\(\left(5^{128}+1\right)\)

=\(12\left(5^{64}-1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)\)

=12.\(\left(5^{128}-1\right)\)\(\left(5^{128}+1\right)\)=12.(5256 - 1)

\(\Rightarrow\)A=\(\dfrac{12\left(5^{256}-1\right)}{5^2-1}\)=\(\dfrac{12\left(5^{256}-1\right)}{24}=\dfrac{5^{256}-1}{2}\)<B=5256-1

Vậy A<B.Chúc các bn học tốtbanh

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

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

9 tháng 8 2016

a) \(A=1999\cdot2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)

=> \(A< B\)

b) \(A=12^6\)

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

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

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

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

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

=> \(A>B\)

c) \(A=2011\cdot2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)

   \(B=2012^2\)

=> \(A< B\)

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

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

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

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

           \(=\frac{3^{128}-1}{2}\)

 \(B=3^{128}-1\)

=> \(A< B\)

Cảm ơn bạn 

B=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1

Vậy B<A

27 tháng 8 2017

a) Ta có : (x + 5)2 - 16 = 0

=> (x + 5)2 = 16

=> (x + 5)2 = (-4) ; 4

\(\Leftrightarrow\orbr{\begin{cases}x+5=-4\\x+5=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=-1\end{cases}}\)