Đáp án:
\(a)\,A=\left\{ {1;\dfrac{3}{2};3}\right\}\\b)\,B=\left\{{-1;0;1;3;7}\right\}\\c)\,C=\left\{{\dfrac{1}{6};1;2;3}\right\}\\d)\,D=\left\{{1;\dfrac{3}{2}}\right\}\)
Giải thích các bước giải:
\(a)\,(2x^2-5x+3)(x^2-4x+3)=0\\\Leftrightarrow \left[ \begin{array}{l}2x^2-5x+3=0\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{3}{2}\\x=1\end{array} \right.\\x^2-4x+3=0\Leftrightarrow\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\end{array} \right.\\\Rightarrow A=\left\{ {1;\dfrac{3}{2};3}\right\}\\b)\,(x^2-10x+21)(x^3-x)=0\\\Leftrightarrow\left[ \begin{array}{l}x^2-10x+21=0\Leftrightarrow\left[ \begin{array}{l}x=7\\x=3\end{array} \right.\\x^3-x=0\Leftrightarrow\left[ \begin{array}{l}x=\pm1\\x=0\end{array} \right.\end{array} \right.\\\Rightarrow B=\left\{{-1;0;1;3;7}\right\}\\c)\,(6x^2-7x+1)(x^2-5x+6)=0\\\Leftrightarrow\left[ \begin{array}{l}6x^2-7x+1=0\Leftrightarrow\left[ \begin{array}{l}x=1\\x=\dfrac{1}{6}\end{array} \right.\\x^2-5x+6=0\Leftrightarrow\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\end{array} \right.\\\Rightarrow C=\left\{{\dfrac{1}{6};1;2;3}\right\}\\d)\,2x^2-5x+3=0\Leftrightarrow\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=1\end{array} \right.\\\Rightarrow D=\left\{{1;\dfrac{3}{2}}\right\}\)
