Đáp án:
$\begin{array}{l}
Dkxd:a \ge 0\\
\dfrac{1}{{4a - 2\sqrt a + 1}} = \dfrac{{3{b^2} + 4}}{3}\\
\Rightarrow \left( {4a - 2\sqrt a + 1} \right).\left( {3{b^2} + 4} \right) = 3\\
Có:4a - 2\sqrt a + 1\\
= {\left( {2\sqrt a } \right)^2} - 2.2\sqrt a .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {2\sqrt a - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
+ )3{b^2} + 4 \ge 4\\
\Rightarrow \left( {4a - 2\sqrt a + 1} \right).\left( {3{b^2} + 4} \right) \ge \dfrac{3}{4}.4 = 3\\
\Rightarrow Dau\, = \,xay\,ra:\\
\Rightarrow \left\{ \begin{array}{l}
2\sqrt a - \dfrac{1}{2} = 0\\
b = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt a = \dfrac{1}{4}\\
b = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \dfrac{1}{{16}}\\
b = 0
\end{array} \right.
\end{array}$
