Đáp án:
Giải thích các bước giải:
$ BPT\Leftrightarrow {{x}^{2}}+{{a}^{2}}\le x\sqrt{{{x}^{2}}+{{a}^{2}}}+2{{\text{a}}^{2}}\Leftrightarrow {{x}^{2}}-{{a}^{2}}\le x\sqrt{{{x}^{2}}+{{a}^{2}}}\left( * \right) $
+ Nếu $ x\ge 0\Rightarrow \left( * \right)\Leftrightarrow {{x}^{2}}-{{a}^{2}}\le \sqrt{{{x}^{2}}\left( {{x}^{2}}+{{a}^{2}} \right)} $
$ \Leftrightarrow \left[ \begin{array}{l} {{x}^{2}}-{{a}^{2}}\le 0 \\ \left\{ \begin{array}{l} {{x}^{2}}-{{a}^{2}}\ge 0 \\ {{\left( {{x}^{2}}-{{a}^{2}} \right)}^{2}}\le {{x}^{2}}\left( {{x}^{2}}+{{a}^{2}} \right) \end{array} \right. \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} \left| x \right|\le \left| a \right| \\ \left\{ \begin{array}{l} \left| x \right|\ge \left| a \right| \\ \left| x \right|\ge \dfrac{\left| a \right|}{\sqrt{3}} \end{array} \right. \end{array} \right.\Leftrightarrow x\ge 0 $
+ Nếu $ x < 0 $ thì BPT:
$ \Leftrightarrow \left\{ \begin{array}{l} {{x}^{2}}-{{a}^{2}}\le 0 \\ {{({{x}^{2}}-{{a}^{2}})}^{2}}\ge {{x}^{2}}({{x}^{2}}+{{a}^{2}}) \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} -\left| a \right|\le x\le \left| a \right| \\ -\dfrac{\left| a \right|}{\sqrt{3}}\le x\le \dfrac{\left| a \right|}{\sqrt{3}} \end{array} \right.\Leftrightarrow -\dfrac{\left| a \right|}{\sqrt{3}}\le x < 0 $
Vậy nghiệm của BPT là $ x\ge -\dfrac{\left| a \right|}{\sqrt{3}} $
