$\begin{array}{l}
A = \dfrac{{{x^2} - 10x + 25}}{{{x^2} - 5x}}\\
a)\,\,DKXD:\,\,\,\\
{x^2} - 5x \ne 0 \Leftrightarrow x\left( {x - 5} \right) \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 5
\end{array} \right.\\
b)\,\,A = \dfrac{{{x^2} - 10x + 25}}{{{x^2} - 5x}}\\
A = \dfrac{{{{\left( {x - 5} \right)}^2}}}{{x\left( {x - 5} \right)}} = \dfrac{{x - 5}}{x}\\
d)\,\,Thay\,\,x = 1\,\,vao\,\,bieu\,\,thuc\,\,A\,\,ta\,\,co\,\,:\\
A = \dfrac{{1 - 5}}{1} = \dfrac{{ - 4}}{1} = - 4
\end{array}$