Đáp án đúng: D
Giải chi tiết:Ta có:
\(\begin{array}{l}\,\,\,\,\,\,\dfrac{{{x^2} + 3}}{2}{\log _2}x = 1 - {\log _2}\dfrac{y}{x} - {x^2}{\log _{\frac{1}{2}}}\sqrt x \\ \Leftrightarrow \dfrac{{{x^2} + 3}}{2}{\log _2}x = 1 - {\log _2}\dfrac{y}{x} + \dfrac{{{x^2}}}{2}{\log _2}x\\ \Leftrightarrow \dfrac{3}{2}{\log _2}x = 1 - {\log _2}y + {\log _2}x\\ \Leftrightarrow \dfrac{1}{2}{\log _2}x = {\log _2}\dfrac{2}{y}\\ \Leftrightarrow {\log _2}\sqrt x = {\log _2}\dfrac{2}{y}\\ \Leftrightarrow \sqrt x = \dfrac{2}{y} \Leftrightarrow y = \dfrac{2}{{\sqrt x }}\\ \Leftrightarrow {y^3} = \dfrac{8}{{x\sqrt x }}\end{array}\)
Khi đó ta có \(P = 4{x^3} + \dfrac{8}{{x\sqrt x }} - 4{\log _2}\left( {4{x^3} + \dfrac{8}{{x\sqrt x }}} \right)\).
Đặt \(t = 4{x^3} + \dfrac{8}{{x\sqrt x }}\), ta có
\(\begin{array}{l}t'\left( x \right) = 12{x^2} - \dfrac{{8.\dfrac{3}{2}{x^{\frac{1}{2}}}}}{{{x^3}}} = 12{x^2} - \dfrac{{12\sqrt x }}{{{x^3}}}\\t'\left( x \right) = 12{x^5} - 12\sqrt x = 0 \Leftrightarrow {x^5} = \sqrt x \\\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {x^{10}} = x \Leftrightarrow x\left( {{x^9} - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\end{array}\)
BBT:
Từ BBT ta suy ra \(t \ge 12\).
Ta có \(P = t - 4{\log _2}t\) với \(t \ge 12\). Ta có \(P' = 1 - \dfrac{4}{{t\ln 2}} = 0 \Leftrightarrow t\ln 2 = 4 \Leftrightarrow t = \dfrac{4}{{\ln 2}} = {t_0} \approx 5,77
otin \left[ {12; + \infty } \right)\).
\( \Rightarrow \) Hàm số đồng biến trên \(\left[ {12; + \infty } \right)\).
\(\begin{array}{l} \Rightarrow \min P = P\left( {12} \right) = 12 - 4{\log _2}12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12 - 4{\log _2}\left( {{2^2}.3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12 - 4\left( {2 + {{\log }_2}3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4 - 4{\log _2}3\end{array}\)
\( \Rightarrow a = 4,\,\,b = - 4,\,\,c = 3\).
Vậy \({a^2} + {b^2} - c = {4^2} + {\left( { - 4} \right)^2} - 3 = 29\).
Chọn D.
