Ta có $a+b+c+2\sqrt{ac+bc}=c+(a+b)+2\sqrt{c(a+b)}=(\sqrt{c}+\sqrt{a+b})^{2}$Tương tự: $a+b+c-2\sqrt{ac+bc}=(\sqrt{c}-\sqrt{a+b})^{2}$
$\rightarrow A= \sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}=\sqrt{c}+\sqrt{a+b}+\left| {\sqrt{c}-\sqrt{a+b}} \right|$
*Nếu $c>a+b$ thì $A=2\sqrt{c}$
*Nếu $c<a+b$ thì $A=2\sqrt{a+b}$