Lời giải:
Ta có:
\(A^2=(\sqrt{x^2-4x+5}-\sqrt{x^2+6x+13})^2=2x^2+2x+18-2\sqrt{(x^2-4x+5)(x^2+6x+13)}(*)\)
Áp dụng BĐT Bunhiacopxky:
\((x^2-4x+5)(x^2+6x+13)=[(x-2)^2+1^2][(x+3)^2+2^2]\)
\(\geq [(x-2)(x+3)+1.2]^2=(x^2+x-4)^2\)
\(\Rightarrow \sqrt{(x^2-4x+5)(x^2+6x+13)}\geq |x^2+x-4|\geq x^2+x-4(**)\)
Từ \((*); (**)\Rightarrow A^2\leq 2x^2+2x+18-2(x^2+x-4)\)
\(\Leftrightarrow A^2\leq 26\Rightarrow A\leq \sqrt{26}\)
Vậy $A_{\max}=\sqrt{26}$. Dấu "=" xảy ra khi $x=7$