a) $x^2-10x+21x^3-x=0$ $\Rightarrow 21x^3+x^2-11x=0$ $\Rightarrow x(21x^2+x-11)=0$ $\Rightarrow \left[ \begin{array}{l}x=0 \\ 21x^2+x-11(1) \end{array} \right .$ Giải $(1)$: $\Delta =1^2-4.21.(-11)=925$ $\Rightarrow \left[ \begin{array}{l}x_1=\dfrac{-1-5\sqrt{37}}{42} \\ x_2=\dfrac{-1+5\sqrt{37}}{42} \end{array} \right .$ $\Rightarrow B=\{0;\dfrac{-1\pm5\sqrt{37}}{42}\}$ b) $\dfrac{1}{2^a}\ge\dfrac{1}{32}=\dfrac{1}{2^5}$ $\Rightarrow a\le5$ $\Rightarrow a=\{0;1;2;3;4;5\}$ $\Rightarrow x=\{1;\dfrac{1}{2};\dfrac{1}{2^2};\dfrac{1}{2^3};\dfrac{1}{2^4};\dfrac{1}{2^5}\}$ $\Rightarrow C=\{1;\dfrac{1}{2};\dfrac{1}{4};\dfrac{1}{8};\dfrac{1}{16};\dfrac{1}{32}\}$

\[\begin{array}{l} B = \left\{ {x \in R|\,\,{x^2} - 10x + 21{x^3} - x = 0} \right\}\\ {x^2} - 10x + 21{x^3} - x = 0\\ \Leftrightarrow 21{x^3} + {x^2} - 11x = 0\\ \Leftrightarrow x\left( {21{x^2} + x - 11} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \frac{{ - 1 + 5\sqrt {37} }}{{42}}\\ x = \frac{{ - 1 - 5\sqrt {37} }}{{42}} \end{array} \right.\\ \Rightarrow B = \left\{ {0;\,\,\frac{{ - 1 - 5\sqrt {37} }}{{42}};\,\,\frac{{ - 1 + 5\sqrt {37} }}{{42}}} \right\}.\\ C = \left\{ {x \in Q|\,\,x = \frac{1}{{{2^a}}},\,\,x \ge \frac{1}{{32}},\,\,a \in N} \right\}\\ x \ge \frac{1}{{32}} \Leftrightarrow \frac{1}{{{2^a}}} \ge \frac{1}{{{2^5}}} \Leftrightarrow a \le 5\\ Lai\,\,co\,\,a \in N\\ \Rightarrow a \in \left\{ {0;\,\,1;\,\,2;\,\,3;\,\,4;\,\,5} \right\}\\ \Rightarrow C = \left\{ {\frac{1}{{{2^5}}};\,\,\frac{1}{{{2^4}}};\,\,\frac{1}{{{2^3}}};\,\,\frac{1}{{{2^2}}};\,\,\frac{1}{2};\,\,1} \right\} = \left\{ {\frac{1}{{32}};\,\,\frac{1}{{16}};\,\,\frac{1}{8};\,\,\frac{1}{4};\,\,\frac{1}{2};\,\,1} \right\}. \end{array}\]