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\(=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)}{2}-\dfrac{3^{16}}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{2}\)
=-1/2
B=(2+1)(22+1)(24+1)(28+1)(216+1)−232
=1.(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(2-1)(2+1)(22+1)(24+1)(28+1)(216+1)−232
=(22-1)(22+1)(24+1)(28+1)(216+1)−232
=(24-1)(24+1)(28+1)(216+1)−232
=(28-1)(28+1)(216+1)−232
=(216-1)(216+1)−232
=232-1-232
=-1
A = ( 2 +1 )( 2^2 + 1 )...(2^16+1) - 2^32
A = ( 2 - 1) ( 2 + 1 )(2^2 + 1) .... (2^16 + 1) - 2^32
A = (2^2 - 1) (2^2 + 1) ...(2^16 + 1) - 2^32
A =( 2^ 4 - 1)( 2^4 + 1 )( 2^8 + 1) (2^16+1) -2^32
A = ( 2^8 - 1)( 2^ 8 + 1) ( 2^ 16 + 1)- 2^32
A = ( 2^16 - 1 )( 2^16 + 1) - 2^32
A = 2^32 - 1 - 2^32
A = - 1
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow2^{64}-1\)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
ta có: 3=2^2-1
thay vào ta được: ''(2^2-1)(2^2+1)'' sử dụng hằng đẳng thức ta được (2^4-1)(2^4+1) tương tự ... ta được đáp án là (2^32-1)
hãy chon đúng cho mình ^ - ^