Đặt \(\sqrt{x+2}=t\in\left[1;\sqrt{3}\right]\)
\(f\left(t\right)=\left|\dfrac{2m\left(t^2-2\right)-4t}{t^2}\right|=\left|-\dfrac{4m}{t^2}-\dfrac{4}{t}+2m\right|=\left|\dfrac{4m}{t^2}+\dfrac{4}{t}-2m\right|\)
Do \(\min\limits_{\left[-1;1\right]}f\left(x\right)=\min\limits_{\left[1;\sqrt{3}\right]}f\left(t\right)>0\)
\(\Rightarrow g\left(t\right)=\dfrac{4m}{t^2}+\dfrac{4}{t}-2m\) vô nghiệm trên \(\left[1;\sqrt{3}\right]\Rightarrow-2< m< 2\sqrt{3}\)
\(\Rightarrow\dfrac{4m}{t^2}+\dfrac{4}{t}-2m>0\Rightarrow\min\limits_{\left[1;\sqrt{3}\right]}f\left(t\right)=f\left(\sqrt{3}\right)=\dfrac{4}{\sqrt{3}}-\dfrac{2m}{3}\)
\(\Rightarrow0< \dfrac{4}{\sqrt{3}}-\dfrac{2m}{3}< 1\Rightarrow m=2\)
Có đúng 1 giá trị nguyên dương m thỏa mãn (4 đáp án đều sai)
