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15 tháng 7 2016

\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)

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

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

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

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

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

\(\Leftrightarrow2^{64}-1\)

10 tháng 8 2023

7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)

\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)

\(A=-\left(1+2+3+...+2004\right)+2005^2\)

\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)

\(A=-1002.2005+2005^2\)

\(A=2005\left(2005-1002\right)=2005.1003=2011015\)

10 tháng 8 2023

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

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

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

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

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

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

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

\(B=-1\)

27 tháng 12 2016

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

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)

26 tháng 10 2023

$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$

$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$

$=\dfrac{3^{32}-1}{2}$

7 tháng 7 2015

đặt A=(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)

=>2A=2.(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)

=>2A=(3-1)(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)

=(32-1)(32+1)(34+1)(38+1)(316+1)(332+1)

=(34-1)(34+1)(38+1)(316+1)(332+1)

=(38-1)(38+1)(316+1)(332+1)

=(316-1)(316+1)(332+1)

=(332-1)(332+1)

=364-1

=>2A=\(\frac{3^{64}-1}{2}\)

15 tháng 7 2018

bạn ơi: A=\(\frac{3^{64}-1}{2}\) chứ ko phải 2A=\(\frac{3^{64}-1}{2}\)

4 tháng 10 2020

Đặt A = ( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

=> 2A = 2.( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

          = ( 3 - 1 )( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

          = ( 32 - 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

          = ( 34 - 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

          = ( 38 - 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )

          = ( 316 - 1 )( 316 + 1 )( 332 + 1 )

          = ( 332 - 1 )( 332 + 1 )

          = 364 - 1

2A = 364 - 1 => A = \(\frac{3^{64}-1}{2}\)