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24 tháng 7 2017
a, \(A=-1^2+2^2-3^2+4^2-...-2017^2+2018^2\)
\(=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)
\(=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)
\(=1+2+3+4+...+2017+2018\)
\(=\dfrac{\left(2018+1\right).2018}{2}=2037171\)
Vậy A=2037171
b, \(B=1^2-2^2+3^2-4^2+...-2004^2+2005^2\)
\(=-\left[\left(2^2-1^2\right)+\left(4^2-3^2\right)+...\left(2004^2-2003^2\right)\right]+2005^2\)
\(=-\left[\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2003+2004\right)\left(2004-2003\right)\right]+2005^2\)
\(=-\left(1+2+3+4+...+2004\right)+2005^2\)
\(=-\dfrac{2005.2004}{2}+2005^2=-2009010+4020025\)
\(=2011015\). Vậy B=2011015
c, \(C=\left(2+1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)...\left(2^{128}+1\right)\)\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{128}+1\right)\)
...
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)=2^{256}-1\)
Vậy \(C=2^{256}-1\)
d, \(D=\left(5+1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(\Rightarrow4D=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(=\left(5^2-1\right)\left(5^2+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
\(=\left(5^4-1\right)\left(5^4+1\right)...\left(5^{2004}+1\right)-5^{2008}\)
...
\(=\left(5^{2004}-1\right)\left(5^{2004}+1\right)-5^{2008}\)
\(=5^{4008}-1-5^{2008}\Rightarrow D=\dfrac{5^{4008}-5^{2008}-1}{4}\)
Vậy \(D=\dfrac{5^{4008}-5^{2004}-1}{4}\)
N
1
BM
21 tháng 12 2017
Ta có: \(K=1^2-2^2+3^2-4^2+......+2005^2\)
\(\Rightarrow K=1^2+\left(3^2-2^2\right)+\left(5^2-4^2\right)+.....\) \(+\left(2005^2-2004^2\right)\)
\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)\)\(+......+\left(2005-2004\right)\left(2005+2004\right)\)
\(\Rightarrow K=1+5+9+13+.....+4009\)
Số số hạng trong tổng K là \(\frac{4009-1}{4}+1=1003\)
\(\Rightarrow K=\frac{\left(4009+1\right).1003}{2}=2005.1003\) = 2011015
KT
13 tháng 2 2018
a) \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\)
\(\Leftrightarrow\)\(\frac{x+1}{4}-1-\frac{x+2}{5}+1+\frac{x+4}{7}-1-\frac{x+5}{8}+1+\frac{x+7}{10}-1-\frac{x+9}{12}+1=0\)
\(\Leftrightarrow\)\(\frac{x-3}{4}-\frac{3-x}{5}+\frac{x-3}{7}-\frac{3-x}{8}+\frac{x+3}{10}-\frac{3-x}{12}=0\)
\(\Leftrightarrow\)\(\frac{x-3}{4}+\frac{x-3}{5}+\frac{x-3}{7}+\frac{x-3}{8}+\frac{x-3}{10}+\frac{x-3}{12}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=0\)
Vì \(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\ne0\)
\(\Rightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Vậy...
b) \(\frac{x}{2004}+\frac{x+1}{2005}+\frac{x+2}{2006}+\frac{x+3}{2007}=4\)
\(\Leftrightarrow\)\(\frac{x}{2004}-1+\frac{x+1}{2005}-1+\frac{x+2}{2006}-1+\frac{x+3}{2007}-1=0\)
\(\Leftrightarrow\)\(\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\)
\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\)
\(\Rightarrow\)\(x-2004=0\)
\(\Leftrightarrow\)\(x=2004\)
Vậy...