b=(1-1/2) x (1-1/3)x (1-1/4 )... (1-1/2003x(1-1/2004) = ?
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TT
1
15 tháng 4 2017
1, \(x^2-2003x-2004=0\)
\(\Rightarrow x^2+x-\left(2004x+2004\right)=0\)
\(\Rightarrow x\left(x+1\right)-2004\left(x+1\right)=0\)
\(\Rightarrow\left(x-2004\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2004=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2004\\x=-1\end{matrix}\right.\)
Vậy x = 2004 hoặc x = -1
2, \(2005x^2-2004x-1=0\)
\(\Rightarrow2005x^2-2005x+x-1=0\)
\(\Rightarrow2005x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(2005x+1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2005x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2005}\\x=1\end{matrix}\right.\)
Vậy \(x=\dfrac{-1}{2005}\) hoặc x = 1
KV
14 tháng 6 2023
a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)
\(=\dfrac{1}{2004}\)
b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(=\dfrac{59}{15}-\dfrac{10}{3}\)
\(=\dfrac{3}{5}\)
\(b=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(b=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2002}{2003}.\frac{2003}{2004}\)
\(b=\frac{1.2.3....2002.2003}{2.3.4....2003.2004}\)
\(b=\frac{1}{2004}\)