Đáp án:
$\begin{array}{l}
2014\left| {x - 12} \right| + {(x - 12)^2} = 2013\left| {12 - x} \right|\\
\Rightarrow 2014\left| {x - 12} \right| - 2013\left| {x - 12} \right| + {(x - 12)^2} = 0\\
\Rightarrow \left| {x - 12} \right| + {(x - 12)^2} = 0\\
Do:\left\{ \begin{array}{l}
\left| {x - 12} \right| \ge 0\\
{\left( {x - 12} \right)^2} \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left| {x - 12} \right| = 0\\
{\left( {x - 12} \right)^2} = 0
\end{array} \right.\\
\Rightarrow x = 12\\
Vậy\,x = 12\\
b)\left\{ \begin{array}{l}
x.\left( {x - y} \right) = \dfrac{3}{{10}}\\
y.\left( {x - y} \right) = \dfrac{{ - 3}}{{50}}
\end{array} \right.\\
\Rightarrow \dfrac{{x\left( {x - y} \right)}}{{y\left( {x - y} \right)}} = \dfrac{3}{{10}}:\dfrac{{ - 3}}{{50}} = 5\\
\Rightarrow \dfrac{x}{y} = 5\\
\Rightarrow x = 5.y\\
\Rightarrow 5.y\left( {5.y - y} \right) = \dfrac{3}{{10}}\\
\Rightarrow 5y.4y = \dfrac{3}{{10}}\\
\Rightarrow {y^2} = \dfrac{3}{{200}}\\
\Rightarrow y = \dfrac{{\sqrt 3 }}{{10\sqrt 2 }} = \dfrac{{\sqrt 6 }}{{20}}\\
\Rightarrow x = \dfrac{{\sqrt 6 }}{4}
\end{array}$
