Đáp án:
đặt $\left\{ \begin{array}{l}
a = \frac{{{d^2}}}{{4{x^2}}}\\
b = 1 - \frac{{{d^2}}}{{4{x^2}}}
\end{array} \right.$
${B_{max}} \Leftrightarrow {\left( {ab} \right)_{max}}$
ta có: $ab \le \frac{{{a^2} + {b^2}}}{2}$
dấu = xảy ra khi a=b $ \Rightarrow \frac{{{d^2}}}{{4{x^2}}} = 1 - \frac{{{d^2}}}{{4{x^2}}}$
nên ${B_{max}} \Leftrightarrow \frac{{{d^2}}}{{4{x^2}}} = 1 - \frac{{{d^2}}}{{4{x^2}}}$
