Đáp án:
$C.{R_3} = 2,4\Omega $
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{12x}}{{12 + x}}\\
{U_{23}} = \dfrac{{{R_{23}}}}{{{R_1} + {R_{23}}}}U\\
\Rightarrow {P_3} = \dfrac{{{U_{23}}^2}}{{{R_3}}} = \dfrac{{{R_{23}}^2}}{{{R_3}.{{\left( {{R_1} + {R_{23}}} \right)}^2}}}.{U^2} = \dfrac{{\dfrac{{144{x^2}}}{{{{\left( {x + 12} \right)}^2}}}}}{{x.{{\left( {3 + \dfrac{{12x}}{{x + 12}}} \right)}^2}}}.{U^2}\\
\Leftrightarrow {P_3} = \dfrac{{144x}}{{{{\left( {15x + 36} \right)}^2}}}{U^2}\\
\Leftrightarrow {P_3} = \dfrac{{144{U^2}}}{{225x + 1080 + \dfrac{{1296}}{x}}}
\end{array}$
Áp dụng bất đẳng thức Cô sy ta có:
$\begin{array}{l}
225x + \dfrac{{1296}}{x} \ge 2.\sqrt {225x.\dfrac{{1296}}{x}} \\
\Leftrightarrow {P_3} \le \dfrac{{144{U^2}}}{{2160}} = \dfrac{{{U^2}}}{{15}}
\end{array}$
Dấu bằng xảy ra khi:
$225x = \dfrac{{1296}}{x} \Rightarrow x = 2,4\Omega $
