Đáp án:
\(\begin{array}{l}
a)\,\,\,6{x^3}y - 9x{y^2}z + 3{x^4}y\\
b)\,\,4{x^3} - 2{x^2} + 1\\
c)\,\,36 + xy\\
d)\,\,4{x^2} - 8y + 5xy\\
e)\,\,x - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,3xy.\left( {2{x^2} - 3yz + {x^3}} \right) = 6{x^3}y - 9x{y^2}z + 3{x^4}y.\\
b)\,\,\left( {24{x^5} - 12{x^4} + 6{x^2}} \right):6{x^2} = 6{x^2}\left( {4{x^3} - 2{x^2} + 1} \right):6{x^2} = 4{x^3} - 2{x^2} + 1.\\
c)\,\,{\left( {2x + 3} \right)^2} + {\left( {2x - 3} \right)^2} - \left( {2x + 3} \right)\left( {4x - 6} \right) + xy\\
= \,{\left( {2x + 3} \right)^2} + {\left( {2x - 3} \right)^2} - 2\left( {2x + 3} \right)\left( {2x - 3} \right) + xy\\
= {\left( {2x + 3 - 2x + 3} \right)^2} + xy = 36 + xy.\\
d)\,\left( {\,8{x^3}y - 16x{y^2} + 10{x^2}{y^2}} \right):\left( {2xy} \right)\\
= 2xy\left( {4{x^2} - 8y + 5xy} \right):\left( {2xy} \right)\\
= 4{x^2} - 8y + 5xy.\\
e)\,\,\left( {{x^3} - 3{x^2} + 3x - 1} \right):\left( {{x^2} - 2x + 1} \right)\\
= {\left( {x - 1} \right)^3}:{\left( {x - 1} \right)^2}\\
= x - 1.
\end{array}\)
