`A={x\inNN|x(x-1)(3x^2-11x-4)=0} \qquad B={x|x=2^k;k\inNN;k<=3}`
`a)` Có: `x(x-1)(3x^2-11x-4)=0`
`<=>x(x-1)(3x^2-12x+x-4)=0`
`<=> x(x-1)[3x(x-4)+(x-4)]=0`
`<=> x(x-1)(x-4)(3x+1)=0`
`<=> [(x=0),(x-1=0),(x-4=0),(3x+1=0):}`
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\x=4\\x=\dfrac{-1}{3}\end{array} \right.\)
Do `x\inNN=>x\in{0;1;4}`
Vậy `A={0;1;4}`
Có: `x=2^k`
Do `k\inNN;k<=3=>k\in{0;1;2;3}`
`=> x\in{1;2;4;8}`
Vậy `B={1;2;4;8}`
`b) A∩B={1;4}`
`\qquad A∪B={0;1;2;4;8}`
`\qquad A\\B={0}`
`\qquad B\\A={2;8}`
