Đáp án + Giải thích các bước giải:
`a//x^{3}-5x^{2}+6x=0`
`⇔(x^{3}-3x^{2})-(2x^{2}-6x)=0`
`⇔x^{2}(x-3)-2x(x-3)=0`
`⇔(x-3)(x^{2}-2x)=0`
`⇔x(x-3)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x-3=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=3\\x=2\end{array} \right.\)
Vậy `S={0;3;2}`
`b//2x^{3}+3x^{2}-32x=48`
`⇔2x^{3}+3x^{2}-32x-48=0`
`⇔(2x^{3}-8x^{2})+(11x^{2}-44x)+(12x-48)=0`
`⇔2x^{2}(x-4)+11x(x-4)+12(x-4)=0`
`⇔(x-4)(2x^{2}+11x+12)=0`
`⇔(x-4)[(2x^{2}+8x)+(3x+12)]=0`
`⇔(x-4)[2x(x+4)+3(x+4)]=0`
`⇔(x-4)(x+4)(2x+3)=0`
`⇔` \(\left[ \begin{array}{l}x-4=0\\x+4=0\\2x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=4\\x=-4\\x=-\frac{3}{2}\end{array} \right.\)
Vậy `S={±4;-(3)/(2)}`
