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ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
ND
Nguyễn Đức Trí
VIP
10 tháng 8 2023
8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
15 tháng 7 2016
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow2^{64}-1\)
11 tháng 7 2015
3(22+1).(24+1).(28+1)......(232+1)+2
=(22-1)(22+1).(24+1).(28+1)......(232+1)+2
=(24-1).(24+1).(28+1)......(232+1)+2
=(28-1).(28+1)......(232+1)+2
=....
=(232-1)(232+1)+2
=264-1+2
=264+1
11 tháng 7 2015
3 = 4 -1 = 22 -1 thay vào ta có :
\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{32}+1\right)+2=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{32}+1\right)+2\)
= \(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+2=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)+2\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)+2=2^{64}-1+2=2^{64}+1\)
CN
1
21 tháng 6 2023
\(=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)}{2}-\dfrac{3^{16}}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{2}\)
=-1/2
LM
1
30 tháng 11 2022
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
MH
2
20 tháng 6 2017
Ta có : $3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)$
$=(2^{16}-1)(2^{16}+1)(2^{32}+1)$
$=(2^{32}-1)(2^{32}+1)$
$=2^{64}-1$
1 tháng 8 2016
VT=[(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)]/2
=[(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)]/2
=[(2^4-1(2^4+1)(2^8+1)(2^16+1)]/2
=[(2^8-1)(2^8+1)(2^16+1)]/2
=[(2^16-1)(2^16+1)]/2
=(2^32-1)/2
\(3.\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
=\(\left(2^{32}-1\right)\left(2^{32}+1\right)\)
=\(2^{64}-1\)