1. $(x + 2)(4 – x) = x^2 + 4x + 4 $
⇔ $(x+2)(4-x)=(x+2)^2$
⇔ $(x+2)(4-x)-(x+2)^2=0$
⇔ $(x+2)(4-x-x-2)=0$
⇔ $(x+2)(-2x+2)=0$
⇔ \(\left[ \begin{array}{l}x+2=0\\-2x+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=1\end{array} \right.\)
Vậy $S=${$-2;1$}
2. $x^2-x-12=0$
⇔ $x^2+3x-4x-12=0$
⇔ $x(x+3)-4(x+3)=0$
⇔ $(x+3)(x-4)=0$
⇔ \(\left[ \begin{array}{l}x+3=0\\x-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-3\\x=4\end{array} \right.\)
3. $ 4x^2 – 12x + 5 = 0$
⇔ $4x^2-10x-2x+5=0$
⇔ $(4x^2-10x)-(2x-5)=0$
⇔ $2x(2x-5)-(2x-5)=0$
⇔ $(2x-5)(2x-1)=0$
⇔ \(\left[ \begin{array}{l}2x-5=0\\2x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{5}{2}\\x=\frac{1}{2}\end{array} \right.\)
Vậy $S=${$\frac{1}{2};\frac{5}{2}$}
