Bạn tham khảo cách chứng minh sau đây nhé:
Ta có:
$\begin{array}{l}
{I_{12}} = \dfrac{U}{{{R_1} + {R_2}}}\\
{I_{34}} = \dfrac{U}{{{R_3} + {R_4}}}
\end{array}$
Vì $U_{MN} = 0$ nên:
$\begin{array}{l}
{U_1} = {U_3}\\
\Leftrightarrow {I_{12}}{R_1} = {I_{34}}{R_3}\\
\Leftrightarrow \dfrac{{U{R_1}}}{{{R_1} + {R_2}}} = \dfrac{{U{R_3}}}{{{R_3} + {R_4}}}\\
\Leftrightarrow {R_1}{R_3} + {R_1}{R_4} = {R_3}{R_1} + {R_3}{R_2}\\
\Leftrightarrow {R_1}{R_4} = {R_2}{R_3}\\
\Leftrightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_3}}}{{{R_4}}}\left( {dpcm} \right)
\end{array}$
