`a) A={x∈RR|(x^2-2x)(2x^2-3x-2)=0}`
Ta có:
`\qquad (x^2-2x)(2x^2-3x-2)=0`
`<=> x(x-2)(2x^2-4x+x-2)=0`
`<=> x(x-2)[2x(x-2)+(x-2)]=0`
`<=> x(x-2)^2(2x+1)=0`
`<=> [(x=0),(x-2=0),(2x+1=0):}`
`<=>`\(\left[ \begin{array}{l}x=0\\x=2\\x=\dfrac{-1}{2}\end{array} \right.\)
Vậy `A={0;2;-1/2}`
`b) B={x\inZZ||x|<5}`
Có: `|x|<5`
`<=> -5<x<5`
Do `x\inZZ=>x\in{-4;-3;-2;-1;0;1;2;3;4}`
Vậy `B={-4;-3;-2;-1;0;1;2;3;4}`
`c) C={x\inZZ|3<|x|<=19/2}`
Có: `3<|x|<=19/2`
`<=>`\(\left[ \begin{array}{l}3<x\le\dfrac{19}{2}\\-3>x\ge\dfrac{-19}{2}\end{array} \right.\)
Do `x\inZZ=>x\in{+-4;+-5;+-6;+-7;+-8;+-9}`
Vậy `C={+-4;+-5;+-6;+-7;+-8;+-9}`
`d) D={n\inN**|3<n^2<30}`
Có: `3<n^2<30`
`<=>`\(\left[ \begin{array}{l}\sqrt{3}<x<\sqrt{30}\\-\sqrt{3}>x>-\sqrt{30}\end{array} \right.\)
Do `n\inN**=>n\in{2;3;4;5}`
Vậy `D={2;3;4;5}`
`e) E={x|x=3k; k \in ZZ; -3<x<5}`
Có: `x=3k <=> k=3/x`
Do `k\inZZ=>3 \vdots x`
`=> x∈Ư(3)={-3;-1;1;3}`
Do `-3<x<5 => x\in{-1;1;3}`
Vậy `E={-1;1;3}`
