Đáp án:
Giải thích các bước giải:
a) x³-2x=x²+2
⇔x³-2x-x²+2=0
⇔(x³-2x)-(x²-2)=0
⇔x(x²-2)-(x²-2)=0
⇔(x²-2)(x-1)=0
⇔\(\left[ \begin{array}{l}x²-2=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=±√2\\x=1\end{array} \right.\)
Vậy S=(±√2,1)
b)2x³-5x²-3x=0
⇔x(2x²-5x-3)=0
⇔x(2x²-6x+x-3)=0
⇔x[2x(x-3)+(x-3)]=0
⇔x[(x-3)(2x+1)]=0
⇔\(\left[ x=0 \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
⇔\(\left[ x=0 \begin{array}{l}x=3\\x=-1/2\end{array} \right.\)
Vậy S=(0,,3, -1/2)
