Đáp án:
`a, (-3xy).(x²+2x-1)= -3x³y-6x²y+3xy`
`b, 12x²y^{6}.z³:3x²y²z²`=4y²z
`c, x⁴- x²+2x³-2x`
`=( x⁴- x²)+(2x³-2x)=x²(x²-1)+2x(x²-1)`
`= (x²-1)(x²+2x)=(x-1)(x+1).x(x+2)`
`=x(x-1)(x+1)(x+2)`
`d, (x-1)²-(2x+3)²=0`
`⇒ (x-1-2x-3)(x-1+2x+3)=0`
`⇒(-x-4)(3x+2)=0`
⇒\(\left[ \begin{array}{l}-x-4=0\\3x+2=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-4\\3x=-2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-4\\x=\frac{-2}{3}\end{array} \right.\)
`Vậy x∈{-4;$\frac{-2}{3}$}`