Đáp án:
$\begin{array}{l}
B2)\\
a){u^3} + 4{u^2}t + 4u{t^2} - 25u\\
= u.\left( {{u^2} + 4ut + 4{t^2} - 25} \right)\\
= u.\left[ {{{\left( {u + 2t} \right)}^2} - {5^2}} \right]\\
= u.\left( {u + 2t + 5} \right)\left( {u + 2t - 5} \right)\\
b)2x - 10y - {x^2} + 10xy - 25{y^2}\\
= 2.\left( {x - 5y} \right) - \left( {{x^2} - 10xy + 25{y^2}} \right)\\
= 2\left( {x - 5y} \right) - {\left( {x - 5y} \right)^2}\\
= \left( {x - 5y} \right)\left( {2 - x + 5y} \right)\\
c){x^5} - 8{x^2}\\
= {x^2}\left( {{x^3} - 8} \right)\\
= {x^2}\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)\\
B3)a)2{x^3} - \dfrac{9}{2}x = 0\\
\Leftrightarrow 4{x^3} - 9x = 0\\
\Leftrightarrow x\left( {4{x^2} - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{3}{2}\\
x = - \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{3}{2};x = - \dfrac{3}{2}\\
b){\left( {x - 2} \right)^2} - {\left( {3 - 2x} \right)^2} = 0\\
\Leftrightarrow \left( {x - 2 - 3 + 2x} \right)\left( {x - 2 + 3 - 2x} \right) = 0\\
\Leftrightarrow \left( {3x - 5} \right)\left( {1 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = 1
\end{array} \right.\\
Vậy\,x = 1;x = \dfrac{5}{3}\\
c){x^2}\left( {2x - 1} \right) - 2x + 1 = 0\\
\Leftrightarrow \left( {2x - 1} \right)\left( {{x^2} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = 1\\
x = - 1
\end{array} \right.\\
Vậy\,x = - 1;x = \dfrac{1}{2};x = 1
\end{array}$
