\(\begin{array}{l} A = \left\{ {1;\,\,2} \right\}\\ \Rightarrow {A_1} = \varnothing ;\,\,\,{A_2} = \left\{ 1 \right\};\,\,\,{A_3} = \left\{ 2 \right\};\,\,{A_4} = \left\{ {1;\,\,2} \right\}.\\ B = \left\{ {7;\,\,8;\,\,9} \right\}\\ \Rightarrow {B_1} = \varnothing ;\,\,\,{B_2} = \left\{ 7 \right\};\,\,\,{B_3} = \left\{ 8 \right\};\,\,\,{B_4} = \left\{ 9 \right\};\,\,{B_5} = \left\{ {7;\,\,8} \right\};\,\,{B_6} = \left\{ {8;\,\,9} \right\};\\ {B_8} = \left\{ {7;\,\,9} \right\};\,\,\,{B_9} = \left\{ {7;\,\,8;\,\,9} \right\}.\\ C = \left\{ {x \in R|\,\,3{x^2} - 4x + 1 = 0} \right\}\\ 3{x^2} - 4x + 1 = 0 \Leftrightarrow \left[ \begin{array}{l} x = 1\\ x = \dfrac{1}{3} \end{array} \right. \Rightarrow C = \left\{ {1;\,\,\dfrac{1}{3}} \right\}\\ \Rightarrow {C_1} = \varnothing ;\,\,\,{C_2} = \left\{ 1 \right\};\,\,\,{C_3} = \left\{ {\dfrac{1}{3}} \right\};\,\,\,{C_4} = \left\{ {1;\,\,\dfrac{1}{3}} \right\}.\\ D = \left\{ {x \in Z|\,\,{x^3} + 4{x^2} + 3x = 0} \right\}\\ {x^3} + 4{x^2} + 3x = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - 1\\ x = - 3 \end{array} \right. \Rightarrow D = \left\{ {0;\,\, - 1; - 3} \right\} \end{array}\)
Câu D em làm tương tự câu B nhé.
