Giải bởi Vietjack
Lời giải
Đáp án đúng là: A
Ta có: \[{\rm{1}}\frac{{\rm{1}}}{{\rm{4}}}{{\rm{x}}^{\rm{2}}}{\rm{y}}\left( { - \frac{{\rm{6}}}{{\rm{5}}}{\rm{xy}}} \right)\left( { - {\rm{2}}\frac{{\rm{1}}}{{\rm{3}}}{\rm{xy}}} \right)\]
\[{\rm{ = }}\left[ {\frac{{\rm{5}}}{{\rm{4}}}{\rm{.}}\left( { - \frac{{\rm{6}}}{{\rm{5}}}} \right){\rm{.}}\left( {\frac{{ - {\rm{7}}}}{{\rm{3}}}} \right)} \right]\left( {{{\rm{x}}^{\rm{2}}}{\rm{.x}}{\rm{.x}}} \right){\rm{.}}\left( {{\rm{y}}{\rm{.y}}{\rm{.y}}} \right){\rm{ = }}\frac{{\rm{7}}}{{\rm{2}}}{{\rm{x}}^{\rm{4}}}{{\rm{y}}^{\rm{3}}}{\rm{.}}\]
Cho đa thức \[{\rm{4}}{{\rm{x}}^{\rm{5}}}{{\rm{y}}^{\rm{2}}} - {\rm{5}}{{\rm{x}}^{\rm{3}}}{\rm{y}} + {\rm{7}}{{\rm{x}}^{\rm{3}}}{\rm{y}} + {\rm{2a}}{{\rm{x}}^{\rm{5}}}{{\rm{y}}^{\rm{2}}}\]. Tìm a để bậc đa thức bằng 4.
