Đáp án:
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sqrt {{x^2} - \dfrac{1}{4} + \sqrt {{x^2} + x + \dfrac{1}{4}} } \ge 0,\,\,\forall x\\
\Rightarrow \dfrac{1}{2}.\left( {2{x^3} + {x^2} + 2x + 1} \right) \ge 0\\
\Leftrightarrow 2{x^3} + {x^2} + 2x + 1 \ge 0\\
\Leftrightarrow {x^2}\left( {2x + 1} \right) + \left( {2x + 1} \right) \ge 0\\
\Leftrightarrow \left( {2x + 1} \right)\left( {{x^2} + 1} \right) \ge 0\\
{x^2} + 1 \ge 0,\,\,\,\forall x \Rightarrow 2x + 1 \ge 0 \Rightarrow x \ge - \dfrac{1}{2}\\
\sqrt {{x^2} - \dfrac{1}{4} + \sqrt {{x^2} + x + \dfrac{1}{4}} } = \dfrac{1}{2}.\left( {2{x^3} + {x^2} + 2x + 1} \right)\\
\Leftrightarrow \sqrt {{x^2} - \dfrac{1}{4} + \sqrt {{{\left( {x + \dfrac{1}{2}} \right)}^2}} } = \dfrac{1}{2}.\left( {2x + 1} \right)\left( {{x^2} + 1} \right)\\
\Leftrightarrow \sqrt {{x^2} - \dfrac{1}{4} + \left| {x + \dfrac{1}{2}} \right|} = \dfrac{1}{2}\left( {2x + 1} \right)\left( {{x^2} + 1} \right)\\
\Leftrightarrow \sqrt {{x^2} - \dfrac{1}{4} + x + \dfrac{1}{2}} = \dfrac{1}{2}\left( {2x + 1} \right).\left( {{x^2} + 1} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - \dfrac{1}{2} \Rightarrow \left| {x + \dfrac{1}{2}} \right| = x + \dfrac{1}{2}} \right)\\
\Leftrightarrow \sqrt {{x^2} + x + \dfrac{1}{4}} = \dfrac{1}{2}\left( {2x + 1} \right)\left( {{x^2} + 1} \right)\\
\Leftrightarrow \sqrt {{{\left( {x + \dfrac{1}{2}} \right)}^2}} = \left( {x + \dfrac{1}{2}} \right).\left( {{x^2} + 1} \right)\\
\Leftrightarrow x + \dfrac{1}{2} = \left( {x + \dfrac{1}{2}} \right)\left( {{x^2} + 1} \right)\\
\Leftrightarrow {x^2}.\left( {x + \dfrac{1}{2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{1}{2}
\end{array} \right.
\end{array}\)
