Đáp án:
$\begin{array}{l}
c)\dfrac{{25}}{{{x^2}}} - \dfrac{{10}}{x} + 1\\
= {\left( {\dfrac{5}{x}} \right)^2} - 2.\dfrac{5}{x} + 1\\
= {\left( {\dfrac{5}{x} - 1} \right)^2}\\
d)4{x^4}{y^8} + {\left( {{x^2}{y^4}} \right)^4} + 4\\
= 4{x^4}{y^8} + {x^8}{y^{16}} + 4\\
= {x^8}{y^{16}} + 4{x^4}{y^8} + 4\\
= {\left( {{x^4}{y^8} + 2} \right)^2}\\
e){\left( {2x - y} \right)^2} + 2\left( {2x - y} \right) + 1\\
= {\left( {2x - y + 1} \right)^2}\\
f){\left( {2x - 4y} \right)^2} + 4x - 8y + 1\\
= {\left( {2x - 4y} \right)^2} + 2\left( {2x - 4y} \right) + 1\\
= {\left( {2x - 4y + 1} \right)^2}
\end{array}$
