Đáp án:
$\begin{array}{l}
1)a)\left( { - \dfrac{1}{3}{x^2}} \right)\left( { - 24y} \right).4xy\\
= \left( { - \dfrac{1}{3}} \right).\left( { - 24} \right).4.{x^2}.x.y.y\\
= 32.{x^3}.{y^2}\\
\Rightarrow \left\{ \begin{array}{l}
\text{Bậc}:5\\
\text{Biến}:{x^3}{y^2}\\
\text{Hệ}\,\text{số}:32
\end{array} \right.\\
b)\left( {x{y^2}} \right)\left( { - 2x{y^3}} \right)\\
= - 2{x^2}{y^5}\\
\Rightarrow \left\{ \begin{array}{l}
\text{Bậc}:7\\
\text{Biến}:{x^2}{y^5}\\
\text{Hệ}\,\text{số}: - 2
\end{array} \right.\\
c)\dfrac{1}{5}{x^2}{y^3}z{\left( {\dfrac{1}{2}xyz} \right)^3}\\
= \dfrac{1}{5}{x^2}{y^3}z.\dfrac{1}{8}{x^3}{y^3}{z^3}\\
= \dfrac{1}{{40}}.{x^5}{y^6}{z^4}\\
\Rightarrow \left\{ \begin{array}{l}
\text{Bậc}:15\\
\text{Biến}:{x^5}{y^6}{z^4}\\
\text{Hệ}\,\text{số}:\dfrac{1}{{40}}
\end{array} \right.\\
d)\dfrac{1}{3}abxy{\left( {a.x.{y^2}} \right)^2}\\
= \dfrac{1}{3}a.b.x.y.{a^2}{x^2}{y^4}\\
= \dfrac{1}{3}{a^3}b.{x^3}.{y^5}\\
\Rightarrow \left\{ \begin{array}{l}
\text{Bậc}:8\\
\text{Biến}:{x^3}{y^5}\\
\text{Hệ}\,\text{số}:\dfrac{1}{3}{a^3}b
\end{array} \right.\\
B2)\\
a)A = {\left( { - \dfrac{2}{3}{x^2}y} \right)^3}.\left( { - 1\dfrac{1}{2}} \right).\left( { - xy} \right)\\
= - \dfrac{8}{{27}}.{x^6}.{y^3}.\left( { - \dfrac{3}{2}} \right).\left( { - xy} \right)\\
= - \dfrac{4}{9}.{x^7}.{y^4}\\
Thay:y = - 2\dfrac{1}{4} = - \dfrac{9}{4} = - \dfrac{{{3^2}}}{{{2^2}}}\\
x = \dfrac{2}{{ - 3}} = - \dfrac{2}{3}\\
\Rightarrow A = - \dfrac{4}{9}.{x^7}.{y^4}\\
= - \dfrac{4}{9}.{\left( { - \dfrac{2}{3}} \right)^7}.{\left( { - \dfrac{{{3^2}}}{{{2^2}}}} \right)^4}\\
= - \dfrac{4}{9}.\dfrac{{ - {2^7}}}{{{3^7}}}.\dfrac{{{3^8}}}{{{2^8}}}\\
= \dfrac{{{2^2}{{.2}^7}{{.3}^8}}}{{{3^2}{{.3}^7}{{.2}^8}}}\\
= \dfrac{3}{2}\\
b)B = \left( { - \dfrac{2}{3}} \right).{\left( { - 1\dfrac{1}{2}x{y^2}} \right)^3}.\left( { - xy} \right)\\
= \dfrac{2}{3}.{\left( { - \dfrac{3}{2}x{y^2}} \right)^3}.x.y\\
= - \dfrac{2}{3}.\dfrac{{{3^3}}}{{{2^3}}}.{x^3}{y^6}.x.y\\
= - \dfrac{{{3^2}}}{{{2^2}}}.{x^4}.{y^7}\\
Thay:x = - 2\dfrac{1}{4} = - \dfrac{{{3^2}}}{{{2^2}}};y = \dfrac{2}{{ - 3}} = \dfrac{{ - 2}}{3}\\
\Rightarrow B = - \dfrac{{{3^2}}}{{{2^2}}}.{\left( { - \dfrac{{{3^2}}}{{{2^2}}}} \right)^4}.{\left( { - \dfrac{2}{3}} \right)^7}\\
= \dfrac{{{3^2}}}{{{2^2}}}.\dfrac{{{3^8}}}{{{2^8}}}.\dfrac{{{2^7}}}{{{3^7}}} = \dfrac{{{3^3}}}{{{2^3}}} = \dfrac{{27}}{8}
\end{array}$
