Đáp án:
e) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
b)\dfrac{{2 - x}}{{2001}} - 1 = \dfrac{{1 - x}}{{2002}} + 1 - \dfrac{x}{{2003}} - 1\\
\to \dfrac{{2003 - x}}{{2001}} = \dfrac{{2003 - x}}{{2002}} - \dfrac{{x - 2003}}{{2003}}\\
\to \dfrac{{2003 - x}}{{2001}} = \dfrac{{2003 - x}}{{2002}} + \dfrac{{2003 - x}}{{2003}}\\
\to \left( {2003 - x} \right)\left( {\dfrac{1}{{2001}} - \dfrac{1}{{2002}} - \dfrac{1}{{2003}}} \right) = 0\\
\to 2003 - x = 0\\
\to x = 2003\\
c){x^2}\left( {x - 3} \right) + 4\left( {3 - x} \right) = 0\\
\to {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\
\to \left( {x - 3} \right)\left( {{x^2} - 4} \right) = 0\\
\to \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 2\\
x = - 2
\end{array} \right.\\
d)\left( {x - 2} \right)\left( {x + 2} \right) + {\left( {x - 2} \right)^2} = 0\\
\to \left( {x - 2} \right)\left( {x + 2 + x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
2x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
e){x^2} - 4x + 3 = 0\\
\to \left( {x - 3} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
