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Giải:
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
\(\Leftrightarrow\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}...+\dfrac{1}{3^{2005}}+\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B-\dfrac{1}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow\dfrac{2}{3}B=\dfrac{1}{3}-\dfrac{1}{3^{2006}}\)
\(\Leftrightarrow B=\dfrac{\dfrac{1}{3}-\dfrac{1}{3^{2006}}}{\dfrac{2}{3}}\)
\(\Leftrightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{\dfrac{3^{2005}-1}{3^{2005}}}{2}\)
\(\Leftrightarrow B=\dfrac{3^{2005}-1}{2.3^{2005}}\)
Vậy ...
Chúc bn học tốt!
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 \(D=\dfrac{5^{4008}-5^{2004}-1}{4}\)
Bạn sửa lại đề bài câu 2) nhé ^^
2) \(a+b+c+d=0\Leftrightarrow a+b=-c-d\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-\left[c^3+d^3+3cd\left(c+d\right)\right]\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\)
\(x^4\left(x^n+1\right)-2\left(x^n+1\right)-x^{n-3}\left(x^{n+3}-x^3\right)+2004\)
\(\Leftrightarrow x^{n+4}+x^4-2x^n-2-x^{n-3+\left(n+3\right)}+x^{n-3+3}+2004\)
\(\Leftrightarrow x^{n+4}+x^4-2x^n-2-x^{x+n}+x^n+2004\)
\(\Leftrightarrow x^{n+4}+x^4-x^n+2002-x^{x+n}\)
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{2005^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}=\frac{2004}{2005}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2005^2}< \frac{2004}{2005}\left(\text{đ}pcm\right)\)
a: \(A=\dfrac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}=2005\)
b: \(B=\dfrac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=2004\)
Có \(\frac{a}{b}=\frac{b}{c}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow a=c.k;b=d.k\)
\(\Rightarrow a^2=c^2.k^2;b^2=d^2.k^2\)
Khi đó \(\frac{a^2+c^2}{b^2+d^2}=\frac{c^2.k^2+c^2}{d^2.k^2+d^2}=\frac{c^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{c^2}{d^2}=\frac{a^2}{b^2}\)
A = 4008 + \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2004}\)
= 4008 + \(\dfrac{1}{2.3:2}+\dfrac{1}{3.4:2}+...+\dfrac{1}{2004.2005:2}\)
= 4008 + \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2004.2005}\)
= 4008 + \(2\left(\dfrac{1}{2}-\dfrac{1}{2005}\right)\)
= 4008 + \(\dfrac{2003}{2005}\)
= 4008\(\dfrac{2003}{2005}\)
a ban oi , day la 4008 chia cho (1+1/1+2+1/1+2+3+.......+1/1+2+....+2004)