Đáp án đúng:
Giải chi tiết:a) \(\,\left( { - \frac{1}{3}xy} \right).\left( {9{x^2}y} \right) = \left( { - \frac{1}{3}.9} \right)xy.{x^2}y = - 3.{x^3}{y^2}\)
Hệ số: \(-3\)
Phần biến: \({x^3}{y^2}\)
b)
\(\begin{array}{l}\,5a{x^2}.{\left( {\frac{{ - 2}}{3}{x^3}{y^2}} \right)^2} = 5a{x^2}.{\left( {\frac{{ - 2}}{3}} \right)^2}.{\left( {{x^3}{y^2}} \right)^2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5a{x^2}.\frac{4}{9}.{x^6}{y^4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {5a.\frac{4}{9}} \right){x^8}{y^4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20a}}{9}{x^8}{y^4}\end{array}\)
Hệ số: \(\frac{{20a}}{9}\)
Phần biến: \({x^8}{y^4}\)