Đáp án:
$\begin{array}{l}
p)P = \dfrac{{81}}{4}{x^2}{\left( {\dfrac{2}{3}xy} \right)^3} - 6xy{\left( {\dfrac{1}{2}{y^2}} \right)^2}\\
+ \dfrac{1}{2}{x^7} - \dfrac{2}{3}{x^5}{y^3} + \dfrac{3}{4}x{y^3}{\left( {\dfrac{{ - 2}}{3}y} \right)^2}\\
= \dfrac{{81}}{4}.{x^2}.\dfrac{8}{{27}}.{x^3}.{y^3} - 6xy.\dfrac{1}{4}{y^4}\\
+ \dfrac{1}{2}{x^7} - \dfrac{2}{3}{x^5}{y^3} + \dfrac{3}{4}.x{y^3}.\dfrac{4}{9}{y^2}\\
= 6{x^5}{y^3} - \dfrac{3}{2}x{y^5} + \dfrac{1}{2}{x^7} - \dfrac{2}{3}{x^5}{y^3} + \dfrac{1}{3}x{y^5}\\
= \dfrac{1}{2}{x^7} + \dfrac{{16}}{3}{x^5}{y^3} - \dfrac{7}{2}x{y^5}\\
\Leftrightarrow Bậc:8\\
q)Q = \dfrac{2}{3}{.9^{x + 1}} - \dfrac{3}{5}{.9^x} + {81.9^{x - 1}}\\
= \dfrac{2}{3}{.9^x}.9 - \dfrac{3}{5}{.9^x} + {9.9.9^{x - 1}}\\
= {6.9^x} - \dfrac{3}{5}{.9^x} + {9.9^x}\\
= \dfrac{{72}}{5}{.9^x}\\
\Leftrightarrow Bậc:x
\end{array}$
