Đáp án:
$\begin{array}{l}
B1)\\
A = 3{a^2}b + 4a{b^2} - \dfrac{1}{2}ab - 4a{b^2} - \dfrac{1}{2}ab - 2 - {a^2}b\\
= \left( {3{a^2}b - {a^2}b} \right) + \left( {4a{b^2} - 4a{b^2}} \right) - \dfrac{1}{2}ab - \dfrac{1}{2}ab - 2\\
= 2{a^2}b - ab - 2\\
1)a = - 2;b = \dfrac{1}{2}\\
\Leftrightarrow A = 2{a^2}b - ab - 2\\
= 2.{\left( { - 2} \right)^2}.\dfrac{1}{2} - \left( { - 2} \right).\dfrac{1}{2} - 2\\
= 4 + 1 - 2\\
= 3\\
2)A - C = a.b - 2\\
\Leftrightarrow 2{a^2}b - ab - 2 - C = a.b - 2\\
\Leftrightarrow C = 2{a^2}b - ab - 2 - ab + 2\\
\Leftrightarrow C = 2{a^2}b - 2ab\\
B2)M = {x^2} - 3{x^4} + {x^3} + 3{x^4} - 2{x^3} - {x^2} - 5x + 6\\
= \left( { - 3{x^4} + 3{x^4}} \right) + \left( {{x^3} - 2{x^3}} \right) + \left( {{x^2} - {x^2}} \right) - 5x + 6\\
= - {x^3} - 5x + 6\\
N\left( x \right) = {x^3} - 3{x^2} + 5x - 2\\
1)P\left( x \right) = M\left( x \right) + N\left( x \right)\\
= - {x^3} - 5x + 6 + {x^3} - 3{x^2} + 5x - 2\\
= - 3{x^2} + 4\\
2)Q\left( x \right) = M\left( x \right) - N\left( x \right)\\
= - {x^3} - 5x + 6 - \left( {{x^3} - 3{x^2} + 5x - 2} \right)\\
= - 2{x^3} + 3{x^2} - 10x + 8\\
3)P\left( a \right) = - 8\\
\Leftrightarrow - 3.{a^2} + 4 = - 8\\
\Leftrightarrow 3{a^2} = 12\\
\Leftrightarrow {a^2} = 4\\
\Leftrightarrow a = 2;a = - 2\\
Vậy\,a = 2;a = - 2
\end{array}$
