Tính nhanh: 1^2-2^2+3^2-4^2+....-2004^2+2005^2
Những câu hỏi liên quan
Tính: A=(1*2004+2*2003+...+2004*1)/(1*2+2*3+...+2004*2005)
1.Tính nhanh: \(\frac{2004^3+1}{2004^2-2003}\); \(\frac{2004^3-1}{2004^2+2005}\)
2. Cho a+b+c+d=0. CMR:
a3+b3+c3+d3=3(ac-bd)(b+d)
Bạn sửa lại đề bài câu 2) nhé ^^
2) \(a+b+c+d=0\Leftrightarrow a+b=-c-d\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-\left[c^3+d^3+3cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)
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9 tháng 8 2023 lúc 21:50
bài 6:tính nhanh
7)1^2-2^2+3^2-4^2+....-2004^2+2005^2
8) (2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}
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bài 6:tính nhanh
7)1\(^2\)-2\(^2\)+3\(^2\)-4\(^2\)+....-2004\(^2\)+2005\(^2\)
8) (2+1)(2\(^2\)+1)(2\(^4\)+1)(2\(^8\)+1)(2\(^{16}\)+1)(2\(^{32}\)+1)-2\(^{64}\)
7) \(A=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(A=\left(-1\right)\left(1^{ }+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(2003+2004\right)+2005^2\)
\(A=-\left(1+2+3+...+2004\right)+2005^2\)
\(A=-\dfrac{2004.\left(2004+1\right)}{2}+2005^2\)
\(A=-1002.2005+2005^2\)
\(A=2005\left(2005-1002\right)=2005.1003=2011015\)
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8) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\dfrac{\left(2^2-1\right)}{2-1}\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(B=\left(2^{64}-1\right)-2^{64}\)
\(B=-1\)
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bài 1 :cho S = 1+2+22+23+24+...........+22005
hãy so sánh:S với 5.22004
bài 2:
hãy tính C=3+12+48+.....+3072+12288
Tính nhanh:
\(1^2-2^2+3^2-4^2+.....2002^2+2003^2-2004^2+2005^2\)
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
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\(-\left(2^2-1^2+4^2-3^2+...+2005^2-2004^2\right)\)
\(=-\left(\left(2-1\right)\left(1+2\right)+...+\left(2005-2004\right)\left(2004+2005\right)\right)\)
\(=-\left(1+2+3+...+2004+2005\right)\)
\(=-\frac{2005\left(2005+1\right)}{2}=-2011015\)
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1^2-2^2+3^2-4^2+...-2004^2+2005^2
(1/2003+1/2004-1/2005)/(5/2003+5/2004-5/2005)-(2/2002+2/2003-2/2004)/(3/2002+3/2003-3/2004)
Tính nhanh :
a) \(A=\dfrac{2004^3+1}{2004^2-2003}\)
b) \(B=\dfrac{2005^3-1}{2005^2+2006}\)
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)
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c = 2005/2 + 2005/3+ 2005/4+....+ 2005/2005 , d = 2006 / 1 + 2006 / 2 + 2006 / 3 +....+ 4009 / 2004 tính c-d