`a) A={x\inRR|(x^2-3x+7)^2=(1-6x)^2}`
Có: `(x^2-3x+7)^2=(1-6x)^2`
`<=> (x^2-3x+7)^2-(1-6x)^2=0`
`<=> (x^2-3x+7-1+6x)(x^2-3x+7+1-6x)=0`
`<=> (x^2+3x+6)(x^2-9x+8)=0`
Do `x^2+3x+6=(x+3/2)^2+15/4>0` với `AAx`
`=> x^2-9x+8=0`
`<=> x^2-x-8x+8=0`
`<=> (x-1)(x-8)=0`
`<=> [(x=1),(x=8):}`
Vậy `A={1;8}`
`b) B={x\inQQ|(x-1)[2x^2-(1+2\sqrt{3})x+\sqrt{3}]=0}`
Có: `(x-1)[2x^2-(1+2\sqrt{3})x+\sqrt{3}]=0`
`<=> (x-1)(2x^2-x-2\sqrt{3}x+\sqrt{3})=0`
`<=> (x-1)[x(2x-1)-\sqrt{3}(2x-1)]=0`
`<=> (x-1)(2x-1)(x-\sqrt{3})=0`
`<=> [(x-1=0),(2x-1=0),(x-\sqrt{3}=0):}`
`<=>`\(\left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\\x=\sqrt{3}\end{array} \right.\)
Do `x\inQQ=>x\in{1;1/2}`
Vậy `B={1;1/2}`
