x²+5x+6=0
⇔(x²+2x)+(3x+6)=0
⇔(x+2)(x+3)=0
⇔\(\left[ \begin{array}{l}x+2=0\\x+3=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=-2\\x=-3\end{array} \right.\)
Vậy S={-2;-3}
x²+3x-4=0
⇔(x²-x)+(4x-4)=0
⇔(x-1)(x+4)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x+4=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1\\x=-4\end{array} \right.\)
Vậy S={1;-4}
x²-3x+2=0
⇔(x²-x)-(2x-2)=0
⇔(x-1)(x-2)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy S={1;2}
x²-9x+20=0
⇔(x²-4x)-(5x-20)=0
⇔(x-4)(x-5)=0
⇔\(\left[ \begin{array}{l}x-4=0\\x-5=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=4\\x=5\end{array} \right.\)
Vậy S={4;5}
x²+2x-15=0
⇔(x²-3x)+(5x-15)=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}.
