Tính nhanh: 1^2-2^2+3^2-4^2+....-2004^2+2005^2
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NM
0
10 tháng 7 2016
Bạn sửa lại đề bài câu 2) nhé ^^
2) \(a+b+c+d=0\Leftrightarrow a+b=-c-d\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-\left[c^3+d^3+3cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)
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) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
TD
2
29 tháng 6 2019
Ta có:
12 - 22 + 32 - 42 + ... + 20032 - 20042 + 20052
= 12 + (-22 + 32) + (-42 + 52) + ... + (-20022 + 20032) +(-20042 + 20052)
= 1 + (32 - 22) + (52 - 42) + ... + (20032 - 20022) + (20052 - 20042)
= 1 + (3 + 2)(3 - 2) + (5 + 4)(5 - 4) + .... + (2003 + 2002)(2003 - 2002) + (2005 + 2004)(2005 - 2004)
= 1 + 5.1 + 9.1 + .... + 4005 . 1 + 4009 . 1
= 1 + (5 + 9 + .... + 4005 + 4009)
= 1 + (4009 + 5)[(4009 - 5) : 4 + 1] : 2
= 1 + 4014 . 1002 : 2
= 1 + 2011014
= 2011015
20 tháng 5 2022
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)
\(1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(=1^2-2^2+3^2-4^2+...+2003^2-2004^2+2005^2\)
\(=-\left(2^2-1^2\right)-\left(4^2-3^2\right)-...-\left(2004^2-2003^2\right)+2005^2\)
\(=-\left[\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2004^2-2003^2\right)\right]+2005^2\)
\(=-\left[\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2004-2003\right)\left(2004+2003\right)\right]+2005^2\)
\(=-\left[1+2+3+4+...+2003+2004\right]+2005^2\)
\(=-\dfrac{2004.2005}{2}+2005^2\)
\(=2011015\)
:D