Đáp án:
$\begin{array}{l}
B1)a)\\
\dfrac{1}{2}{x^2}y\left( {2{x^3} - \dfrac{2}{5}x{y^2} - 1} \right)\\
= {x^5}y - \dfrac{1}{5}{x^3}{y^3} - \dfrac{1}{2}{x^2}y\\
b)\left( {{x^2} - 2x + 3} \right)\left( {\dfrac{1}{2}x - 5} \right)\\
= \dfrac{1}{2}{x^3} - 5{x^2} - {x^2} + 10x + \dfrac{3}{2}x - 15\\
= \dfrac{1}{2}{x^3} - 6{x^2} + \dfrac{{23}}{2}x - 15\\
B2)a)\\
{\left( {\dfrac{1}{2}x + 4} \right)^2} = {\left( {\dfrac{1}{2}x} \right)^2} + 2.\dfrac{1}{2}x.4 + {4^2}\\
= \dfrac{1}{4}{x^2} + 4x + 16\\
b){\left( {5x - \dfrac{1}{5}} \right)^3}\\
= {\left( {5x} \right)^3} - 3.{\left( {5x} \right)^2}.\dfrac{1}{5} + 3.5x.\dfrac{1}{{{5^2}}} - \dfrac{1}{{{5^3}}}\\
= 125{x^3} - 15{x^2} + \dfrac{3}{5}x - \dfrac{1}{{125}}\\
c)\left( {5x - 4} \right)\left( {4 + 5x} \right) = {\left( {5x} \right)^2} - {4^2} = 25{x^2} - 16\\
B3)a)\\
5x\left( {x + 1} \right) - 5\left( {x + 1} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow 5x\left( {x + 1} \right) - \left( {x + 1} \right)\left( {5x - 10} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {5x - 5x + 10} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right).10 = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
Vậy\,x = - 1\\
b)\left( {4x + 1} \right)\left( {x - 2} \right) - {\left( {2x - 3} \right)^2} = 4\\
\Leftrightarrow 4{x^2} - 8x + x - 2 - \left( {4{x^2} - 12x + 9} \right) = 4\\
\Leftrightarrow 4{x^2} - 7x - 2 - 4{x^2} + 12x - 9 = 4\\
\Leftrightarrow 5x = 15\\
\Leftrightarrow x = 3\\
Vậy\,x = 3
\end{array}$
