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\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{2004+1}{1-2003}\)\(=\frac{2005}{-2002}\)
\(B=\frac{2005^3-1}{2005^2+2006}\)\(=\frac{2005-1}{1+2006}=\frac{2004}{2007}\)
\(\Rightarrow A>B\)
\(A=\frac{2004^3+1}{2004^2-2003}\)
\(A=\frac{\left(2004+1\right)\left(2004^2-2004+1\right)}{2004^2-2003}\)
\(A=\frac{2005.\left(2004^2-2003\right)}{2004^2-2003}=2005\)
\(B=\frac{2005^3-1}{2005^2+2006}\)
\(B=\frac{\left(2005-1\right)\left(2005^2+2005+1\right)}{2005^2+2006}=\frac{2004.\left(2005^2+2006\right)}{2005^2+2006}=2004\)
Tham khảo nhé~
......................?
mik ko biết
mong bn thông cảm
nha ................
a) 2 +4+6+8+...+2018
= ( 2018+2) x 1009 : 2
= 2020 x 1009 : 2
= 1009 x (2020:2)
= 1009 x 1010
= 1 019 090
b) S = 10 + 102 + 103 + ...+ 10100
=> 10.S = 102 + 103 + 104 +...+ 10101
=> 10.S - S = 10101-10
9.S=10101- 10
\(\Rightarrow S=\frac{10^{101}-10}{9}\)
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5S-S=1-\frac{1}{5^{100}}\)
\(4S=1-\frac{1}{5^{100}}\)
\(S=\frac{1-\frac{1}{5^{100}}}{4}\)
e cx ko nx, e ms hok lp 7 thoy, sang hè ms lp 8! e sr cj nhiều nha!
d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
\(S=\frac{1009}{2020}\)
Giải phương trình chứ chứng minh cái gì
\(\frac{1}{2x-2006}+\frac{1}{3-2007x}+\frac{1}{2006x+2005}=\frac{1}{x+2}\)
\(\Leftrightarrow\left(\frac{1}{2x-2006}-\frac{1}{x+2}\right)+\left(\frac{1}{3-2007x}+\frac{1}{2006x+2005}\right)=0\)
\(\Leftrightarrow\frac{x-2008}{\left(2x-2006\right)\left(x+2\right)}+\frac{x-2008}{\left(3-2007x\right)\left(2006x-2005\right)}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{\left(2x-2006\right)\left(x+2\right)}+\frac{1}{\left(3-2007x\right)\left(2006x-2005\right)}\right)=0\)
\(\Leftrightarrow\left(x-2008\right)\left(2008x-1\right)\left(2005x+2003\right)=0\)
\(\Leftrightarrow x=2008;x=\frac{1}{2008};x=-\frac{2003}{2005}\)
\(A=\left(\frac{x-2}{2x-2}+\frac{3}{2x-2}-\frac{x+3}{2x+2}\right):\left(-1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x-2}{2\left(x-1\right)}+\frac{3}{2\left(x-1\right)}+\frac{-\left(x+3\right)}{2\left(x+1\right)}\right):\left(-\frac{1}{1}+\frac{-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{-1\left(x+1\right)-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{x^2-x^2+x+3x-2x-6+3+3}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x-1-x+3}{x+1}\right)\)
=\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}:\frac{2}{x+1}\)
\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}.\frac{x+1}{2}\)
\(=\frac{x}{2\left(x-1\right)}\)
b,Thayx=2005
\(\Rightarrow A=\frac{2005}{4008}\)
\(3S=3+\frac{1}{3}+...+\frac{1}{3^{2004}}\)
\(3S-S=\left(3+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)
\(2S=3-\frac{1}{3^{2005}}\)
\(2S=\frac{3^{2006-1}}{3^{2005}}\)
\(S=\frac{3^{2006}-1}{3^{2005}.2}\)
S = 1/3 + 1/32 + 1/33 + ... + 1/32005
=> 3S = 1 + 1/3 + 1/32 + ... + 1/32004
=> 3S - S = 1 + 1/3 + 1/32 + ... + 1/32004 - (1/3 + 1/32 + 1/33 + ... + 1/32005)
=> 2S = 1 + 1/3 + 1/32 + ... + 1/32004 - 1/3 - 1/32 - 1/33 - ... - 1/32005
=> 2S = 1 - 1/32005
=> S = \(\frac{\frac{1}{3^{2005}}}{2}\)
=> S = 1/32005.2