Đáp án:
$m<-2+2\sqrt{2}$
Giải thích các bước giải:
$\sqrt{2-x}+\sqrt{2+x}-\sqrt{4-x^2}-m>0 \ \forall \ x \in [-2;2]\\ \Leftrightarrow \sqrt{2-x}+\sqrt{2+x}-\sqrt{4-x^2}>m \ \forall \ x \in [-2;2]\\ \Leftrightarrow f(x)>m \ \forall \ x \in [-2;2]\\ \Leftrightarrow \underset{ [-2;2]}{min} f(x)>m(1)\\ f(x)=\sqrt{2-x}+\sqrt{2+x}-\sqrt{4-x^2}\\ f'(x)=\dfrac{-1}{2\sqrt{2-x}}+\dfrac{1}{2\sqrt{2+x}}+\dfrac{x}{\sqrt{4-x^2}}\\ =\dfrac{-\sqrt{2+x}}{2\sqrt{2-x}\sqrt{2+x}}+\dfrac{\sqrt{2-x}}{2\sqrt{2+x}\sqrt{2-x}}+\dfrac{2x}{2\sqrt{4-x^2}}\\ =\dfrac{-\sqrt{2+x}+\sqrt{2-x}+2x}{2\sqrt{4-x^2}}\\ f'(x)=0 \Leftrightarrow -\sqrt{2+x}+\sqrt{2-x}+2x=0\\ \Leftrightarrow 2x=\sqrt{2+x}-\sqrt{2-x}\\ \Rightarrow 4x^2=(\sqrt{2+x}-\sqrt{2-x})^2\\ \Leftrightarrow 4x^2=2+x+2-x-2\sqrt{(2+x)(2-x)}\\ \Leftrightarrow 4x^2=4-2\sqrt{4-x^2}\\ \Leftrightarrow 2x^2-2=-\sqrt{4-x^2}(2x^2-2 \le 0)\\ \Rightarrow (2x^2-2)^2=4-x^2\\ \Leftrightarrow 4x^4-8x^2+4=4-x^2\\ \Leftrightarrow 4x^4-7x^2=0\\ \Leftrightarrow x^2(4x^2-7)=0\\ \Leftrightarrow \left[\begin{array}{l} x=0\\ x=\pm \dfrac{\sqrt{7}}{2}(L)\end{array} \right.\\ BBT$
\begin{array}{|c|ccccccccc|} \hline x&-2&&0&&2\\\hline y'&&-&0&+&\\\hline &2&&&&2\\y&&\searrow&&\nearrow&\\&&&-2+2\sqrt{2}\\\hline\end{array}
Từ BBT $\Rightarrow (1) \Leftrightarrow m<-2+2\sqrt{2}.$
