Đáp án:
$\begin{array}{l}
a)Dkxd:y \ne 2;x \ne \pm 5\\
D = \dfrac{{{y^2} - y - 2}}{{y - 2}}:\dfrac{{{x^3} - 10{x^2} + 25x}}{{{x^2} - 25}}\\
= \dfrac{{\left( {y - 2} \right)\left( {y + 1} \right)}}{{y - 2}}.\dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{x{{\left( {x - 5} \right)}^2}}}\\
= \left( {y + 1} \right).\dfrac{{x + 5}}{{x\left( {x - 5} \right)}}\\
b){x^2} + \left| {x - 2} \right| + 4{y^2} - 4xy = 0\\
\Rightarrow {x^2} - 4xy + 4{y^2} + \left| {x - 2} \right| = 0\\
\Rightarrow {\left( {x - 2y} \right)^2} + \left| {x - 2} \right| = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x - 2y} \right)^2} \ge 0\\
\left| {x - 2} \right| \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left| {x - 2} \right| = 0\\
{\left( {x - 2y} \right)^2} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
x = 2y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 1
\end{array} \right.\\
\Rightarrow D = \left( {y + 1} \right).\dfrac{{x + 5}}{{x\left( {x - 5} \right)}}\\
= 2.\dfrac{7}{{2.\left( {2 - 5} \right)}}\\
= \dfrac{{ - 7}}{3}
\end{array}$
