`#tnvt`
`a)(8x^2-4x):(-4x)-(x+2)=8`
`<=>-2x+1-x-2=8`
`<=>-3x=9`
`<=>x=-3`
Vậy `x=-3`
`b)(2x^4-3x^3+x^2):(-x^2)+4(x-1)^2=0`
`<=>-2x^2+3x-1+4(x^2-2x+1)=0`
`<=>-2x^2+3x-1+4x^2-8x+4=0`
`<=>2x^2-5x+3=0`
`<=>2x^2-2x-3x+3=0`
`<=>2x(x-1)-3(x-1)=0`
`<=>(x-1)(2x-3)=0`
`<=>`$\left[\begin{matrix} x=1 \\ x=\dfrac{3}{2}\end{matrix}\right.$
Vậy `x\in{1;3/2}`