Đáp án:
$\begin{array}{l}
28)\dfrac{1}{9}{t^2} - \dfrac{4}{{25}}{f^2}\\
= {\left( {\dfrac{1}{3}t} \right)^2} - {\left( {\dfrac{2}{5}f} \right)^2}\\
= \left( {\dfrac{1}{3}t - \dfrac{2}{5}f} \right)\left( {\dfrac{1}{3}t + \dfrac{2}{5}f} \right)\\
29)\dfrac{1}{4}{\left( {2x - 6y} \right)^2} - \dfrac{1}{9}{\left( {9x - 3y} \right)^2}\\
= {\left( {\dfrac{1}{2}.\left( {2x - 6y} \right)} \right)^2} - {\left( {\dfrac{1}{3}.\left( {9x - 3y} \right)} \right)^2}\\
= {\left( {x - 3y} \right)^2} - {\left( {3x - y} \right)^2}\\
= \left( {x - 3y - 3x + y} \right)\left( {x - 3y + 3x - y} \right)\\
= \left( { - 2x - 2y} \right)\left( {4x - 4y} \right)\\
= - 8\left( {x + y} \right)\left( {x - y} \right)\\
30)2{\left( {\sqrt 2 x - 3\sqrt 2 y} \right)^2} - 5{\left( {2\sqrt 5 y - \sqrt 5 x} \right)^2}\\
= {\left( {\sqrt 2 \left( {\sqrt 2 x - 3\sqrt 2 y} \right)} \right)^2} - {\left( {\sqrt 5 \left( {2\sqrt 5 y - \sqrt 5 x} \right)} \right)^2}\\
= {\left( {2x - 6y} \right)^2} - {\left( {10y - 5x} \right)^2}\\
= \left( {2x - 6y - 10y + 5x} \right)\left( {2x - 6y + 10y - 5x} \right)\\
= \left( {7x - 16y} \right)\left( {4y - 3x} \right)\\
31)\\
16{x^4} - 9{y^6}\\
= \left( {4{x^2} - 3{y^3}} \right)\left( {4{x^2} + 3{y^3}} \right)\\
32)\\
{x^6} - {y^3}\\
= \left( {{x^2} - {y^2}} \right)\left( {{x^4} + {x^2}{y^2} + {y^4}} \right)\\
33)20xy + 25{y^2} + 4{x^2}\\
= {\left( {2x} \right)^2} + 2.2x.5y + {\left( {5y} \right)^2}\\
= {\left( {2x + 5y} \right)^2}\\
34)\\
{x^4} + 25{y^4} + 10{x^2}{y^2}\\
= {\left( {{x^2} + 5{y^2}} \right)^2}\\
35)\\
- 49{x^2} + 25{y^2}\\
= {\left( {5y} \right)^2} - {\left( {7x} \right)^2}\\
= \left( {5y - 7x} \right)\left( {5y + 7x} \right)
\end{array}$
