Đáp án:
$\begin{array}{l}
1)a)M = - \dfrac{1}{9}{x^4}{y^3}{\left( {2x{y^2}} \right)^2}\\
= \dfrac{{ - 1}}{9}{x^4}{y^3}.4{x^2}{y^4}\\
= - \dfrac{4}{9}{x^6}{y^7}\\
b)y = \dfrac{x}{{ - 3}}\\
\Leftrightarrow x = - 3y\\
Khi:x + y = 2\\
\Leftrightarrow - 3y + y = 2\\
\Leftrightarrow - 2y = 2\\
\Leftrightarrow y = - 1\\
\Leftrightarrow x = - 3y = 3\\
M = - \dfrac{4}{9}{x^6}{y^7}\\
= \dfrac{{ - 4}}{9}.{\left( 3 \right)^6}.{\left( { - 1} \right)^7}\\
= {4.3^4}\\
= 324\\
B2)\\
a)A\left( x \right) = 2 - 6{x^3} - {x^2} + 10{x^3} - 2\left( {x - 1} \right) - 4{x^2}\\
= 4{x^3} - 5{x^2} - 2x + 4\\
B\left( x \right) = - 5{x^3} - \left( {{x^2} + 1} \right) + 5x + {x^2} - 8x + 3{x^3}\\
= - 2{x^3} - 3x - 1\\
C\left( x \right) = 2x - 3{x^2} - 4 + {x^3}\\
= {x^3} - 3{x^2} + 2x - 4\\
b)\\
A\left( x \right) + B\left( x \right) - C\left( x \right)\\
= 4{x^3} - 5{x^2} - 2x + 4\\
+ \left( { - 2{x^3} - 3x - 1} \right)\\
- \left( {{x^3} - 3{x^2} + 2x - 4} \right)\\
= {x^3} - 2{x^2} - 7x + 7\\
c)\\
P\left( x \right) = C\left( x \right) - {x^3} + 4\\
= {x^3} - 3{x^2} + 2x - 4 - {x^3} + 4\\
= - 3{x^2} + 2x\\
P\left( x \right) = 0\\
\Leftrightarrow - 3{x^2} + 2x = 0\\
\Leftrightarrow x\left( { - 3x + 2} \right) = 0\\
\Leftrightarrow x = 0;x = \dfrac{2}{3}\\
Vậy\,x = 0;x = \dfrac{2}{3}
\end{array}$
