a) x²+2x=15
⇔x²+2x-15=0
⇔x²-3x+5x-15=0
⇔x(x-3)+5(x-3)=0
⇔(x-3)(x+5)=0
⇔\(\left[ \begin{array}{l}x-3=0\\x+5=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
Vậy S={3;-5}
b)2x³-2x²=4x
⇔2x³-2x²-4x=0
⇔2x(x²-x-2)=0
⇔2x(x-2)(x+1)=0
⇔x=0; x=2; x=-1
Vậy S={0;2;-1}
c)x^4-5x³+4x²=0
⇔x²(x²-5x+4)=0
⇔x²(x-1)(x-4)=0
⇔x=0;x=1;x=4
Vậy S={0;1;4}
d)x³+4x²-9x-36=0
⇔x²(x+4)-9(x+4)=0
⇔(x+4)(x-3)(x+3)=0
⇔x=-4;x=±3
Vậy S={-4;±3}