$a$) $A= x^2 + 4x = x(x+4)$
Để $x(x+4) > 0$ thì : $x;x+4$ cùng dấu.
$TH1$. $\left\{\begin{matrix}x > 0 & \\x + 4 > 0 & \end{matrix}\right.$ $⇒$ $x>0$
$TH2$. $\left\{\begin{matrix}x < 0 & \\x + 4 < 0 & \end{matrix}\right.$ $⇒$ $x<-4$
Vậy $x>0;x<-4$.
$b$) $B = (x-3)(x+7)$
Để $B = (x-3)(x+7) > 0$ thì : $x-3;x+7$ cùng dấu.
$TH1$. $\left\{\begin{matrix}x-3 > 0 & \\x + 7 > 0 & \end{matrix}\right.$ $⇒$ $x>3$
$TH2$. $\left\{\begin{matrix}x-3 < 0 & \\x + 7 < 0 & \end{matrix}\right.$ $⇒$ $x<-7$
Vậy $x>3;x<-7$.
$c$) `C = (1/2 - x).(1/3 - x)`
Để `C = (1/2 - x).(1/3 - x)>0` thì : `1/2 - x; 1/3 - x` cùng dấu
$TH1$. $\left\{\begin{matrix}\dfrac{1}{2}-x > 0 & \\\dfrac{1}{3}-x > 0 & \end{matrix}\right.$ $⇒$ $x < \dfrac{1}{3}$
$TH2$. $\left\{\begin{matrix}\dfrac{1}{2}-x < 0 & \\\dfrac{1}{3}-x < 0 & \end{matrix}\right.$ $⇒$ $x ? \dfrac{1}{2}$
Vậy $x>\dfrac{1}{2}; x < \dfrac{1}{3}$.
