a, $A=(x+1).(x^2+2)=0$
$⇒x+1=0$ (vì: x²+2>0)
$⇒x=-1$
b, $(x^2-4).(x+1)=0$
$⇒$\(\left[ \begin{array}{l}x^2-4=0\\x+1=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x^2=4\\x=-1\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=±2\\x=-1\end{array} \right.\)
Vậy x∈{±2;-1}
c, $(2y-4)^2+|x-2|=0$
Vì: $(2y-4)^2≥0∀y;|x-2|≥0∀x$
Dấu "=" xảy ra khi $\left \{ {{2y-4=0} \atop {x-2=0}} \right.$ $⇒\left \{ {{y=2} \atop {x=2}} \right.$
Vậy x=2 và y=2
d, $x^2-5x+6=0$
$⇒x^2-2x-3x+6=0$
$⇒x.(x-2)-3.(x-2)=0$
$⇒(x-2)(x-3)=0$
$⇒$\(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy x=2 hoặc x=3
