a) $x^{2}$- 2003x- 2004= 0
⇔$x^{2}$+ 2004x- x- 2004= 0
⇔x(x+ 2004)- (x+ 2004)= 0
⇔(x- 1)(x+ 2004)= 0
⇔\(\left[ \begin{array}{l}x- 1= 0\\x+ 2004= 0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x= 1\\x= -2004\end{array} \right.\)
Vậy x= 1 hoặc x= -2004
b) $2005x^{2}$- 2004x- 1= 0
⇔$2005x^{2}$+ 2005x- x- 1= 0
⇔2005x(x+1)- (x+1)= 0
⇔(2005x- 1)(x+ 1)= 0
⇔\(\left[ \begin{array}{l}2005x- 1=0\\x+ 1= 0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2005x= 1\\x=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x= \frac{1}{2005} \\x=-1\end{array} \right.\)
Vậy x= $\frac{1}{2005}$ hoặc x= -1
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