Đáp án:
$\begin{array}{l}
c)C = {\left( {\dfrac{x}{2} - y} \right)^3} - 6{\left( {y - \dfrac{x}{2}} \right)^2} - 12.\left( {y - \dfrac{x}{2}} \right) - 8\\
= {\left( {\dfrac{x}{2} - y} \right)^3} - 6{\left( {\dfrac{x}{2} - y} \right)^2} + 12\left( {\dfrac{x}{2} - y} \right) - 8\\
= {\left( {\dfrac{x}{2} - y - 2} \right)^3}\\
= {\left( {\dfrac{4}{2} - 2 - 2} \right)^3}\\
= {\left( { - 2} \right)^3}\\
= - 8\\
d)D = 2\left( {{x^3} + {y^3}} \right) - 3\left( {{x^2} + {y^2}} \right)\\
= 2\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) - 3\left[ {{{\left( {x + y} \right)}^2} - 2xy} \right]\\
= 1.\left( {2{x^2} - 2xy + 2{y^2}} \right) - 3.\left( {1 - 2xy} \right)\left( {do:x + y = 1} \right)\\
= 2{x^2} - 2xy + 2{y^2} - 3 + 6xy\\
= 2{x^2} + 4xy + 2{y^2} - 3\\
= 2\left( {{x^2} + 2xy + {y^2}} \right) - 3\\
= 2{\left( {x + y} \right)^2} - 3\\
= - 1
\end{array}$
