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