Đáp án:
$\begin{array}{l}
a)\left( {2{x^2} - 5x + 1} \right)\left( { - \dfrac{2}{3}xy} \right) + 2xy\left( {x - 1} \right)\\
= \left( { - \dfrac{2}{3}xy} \right).2{x^2} + \left( { - \dfrac{2}{3}xy} \right).\left( { - 5x} \right) - \dfrac{2}{3}xy + 2{x^2}y - 2xy\\
= - \dfrac{4}{3}{x^3}y + \dfrac{{10}}{3}{x^2}y - \dfrac{2}{3}xy + 2{x^2}y - 2xy\\
= - \dfrac{4}{3}{x^3}y + \dfrac{{16}}{3}{x^2}y - \dfrac{8}{3}xy\\
Thay\,x = 1;y = 2\\
\Rightarrow A = - \dfrac{4}{3}{.1^3}.2 + \dfrac{{16}}{3}{.1^2}.2 - \dfrac{8}{3}.1.2\\
= \dfrac{{ - 8}}{3} + \dfrac{{32}}{3} - \dfrac{{16}}{3}\\
= \dfrac{8}{3}\\
b)B = - 2xy\left( {{x^2} - 1} \right) - x\left( {1 - 2xy} \right)\\
= - 2{x^3}y + 2xy - x + 2{x^2}y\\
x = 2;y = - 2\\
\Rightarrow B = - {2.2^3}.\left( { - 2} \right) + 2.2.\left( { - 2} \right) - 2 + {2.2^2}.\left( { - 2} \right)\\
= 32 - 8 - 2 - 16\\
= 6\\
c)C = \left( {\dfrac{1}{2}xy - y + 1} \right).\left( { - 2x} \right) - x\left( {y - 1} \right)\\
= - {x^2}y + 2xy - 2x - xy + x\\
= - {x^2}y + xy - x\\
Thay\,x = \dfrac{1}{2};y = - \dfrac{1}{2}\\
\Rightarrow C = - {\left( {\dfrac{1}{2}} \right)^2}.\left( { - \dfrac{1}{2}} \right) + \dfrac{1}{2}.\left( { - \dfrac{1}{2}} \right) - \dfrac{1}{2}\\
= \dfrac{1}{8} - \dfrac{1}{4} - \dfrac{1}{2}\\
= \dfrac{{ - 5}}{8}\\
d)x.\left( {{x^2}y - \dfrac{1}{2}xy - \dfrac{3}{2}} \right) - y\left( {{x^3} - 1} \right)\\
= {x^3}y - \dfrac{1}{2}{x^2}y - \dfrac{3}{2}x - {x^3}y + y\\
= - \dfrac{1}{2}{x^2}y - \dfrac{3}{2}x + y\\
Thay\,x = - 1;y = 1\\
\Rightarrow D = - \dfrac{1}{2}.{\left( { - 1} \right)^2}.1 - \dfrac{3}{2}.\left( { - 1} \right) + 1\\
= - \dfrac{1}{2} + \dfrac{3}{2} + 1\\
= 1 + 1 = 2
\end{array}$
