50đ Giải hộ mình bài này với ạ

Đáp án:

$\begin{array}{l}
a)16{x^2} + 24x + 9\\
 = {\left( {4x} \right)^2} + 2.4x.3 + {3^2}\\
 = {\left( {4x + 3} \right)^2}\\
b)\dfrac{{{x^2}}}{4} - 3x + 9\\
 = {\left( {\dfrac{x}{2}} \right)^2} - 2.\dfrac{x}{2}.3 + {3^2}\\
 = {\left( {\dfrac{x}{2} - 3} \right)^2}\\
c)4{u^4}{v^8} + {\left( {{u^2}{v^4}} \right)^4} + 4\\
 = {\left( {{u^4}{v^8}} \right)^2} + 4\left( {{u^4}{v^8}} \right) + 4\\
 = {\left( {{u^4}{v^8} + 2} \right)^2}\\
d)\dfrac{{25}}{{{v^2}}} - \dfrac{{10}}{v} + 1\\
 = {\left( {\dfrac{5}{v}} \right)^2} - 2.\dfrac{5}{v}.1 + 1\\
 = {\left( {\dfrac{5}{v} - 1} \right)^2}\\
e){\left( { - m + 2n} \right)^2} + 2\left( {2n - m} \right) + 1\\
 = {\left( {2n - m} \right)^2} + 2\left( {2n - m} \right).1 + 1\\
 = {\left( {2n - m + 1} \right)^2}\\
f){\left( {2p - 4q} \right)^2} + 4p - 8q + 1\\
 = {\left( {2p - 4q} \right)^2} + 2.\left( {2p - 4q} \right) + 1\\
 = {\left( {2p - 4q + 1} \right)^2}
\end{array}$