Đáp án:
\(\begin{array}{l}
a,\\
A = - 3{x^3} + 6{x^2} + 6\\
x = 5 \Rightarrow A = - 219\\
b,\\
B = - 3{x^3}y + 3{y^3}\\
x = 1;y = \dfrac{1}{2} \Rightarrow B = - \dfrac{9}{8}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = 3\left( {3{x^2} + 2} \right) - {x^2}.\left( {4x + 3} \right) + {x^3}\\
= \left( {9{x^2} + 6} \right) - \left( {4{x^3} + 3{x^2}} \right) + {x^3}\\
= 9{x^2} + 6 - 4{x^3} - 3{x^2} + {x^3}\\
= \left( { - 4{x^3} + {x^3}} \right) + \left( {9{x^2} - 3{x^2}} \right) + 6\\
= - 3{x^3} + 6{x^2} + 6\\
x = 5 \Rightarrow A = - {3.5^3} + {6.5^2} + 6 = - 3.125 + 6.25 + 6 = - 375 + 150 + 6 = - 219\\
b,\\
B = 7xy\left( {xy - {x^2}} \right) - 3{y^2}\left( {{x^2} - y} \right) - 4{x^2}\left( {{y^2} - xy} \right)\\
= \left( {7{x^2}{y^2} - 7{x^3}y} \right) - \left( {3{x^2}{y^2} - 3{y^3}} \right) - \left( {4{x^2}{y^2} - 4{x^3}y} \right)\\
= 7{x^2}{y^2} - 7{x^3}y - 3{x^2}{y^2} + 3{y^3} - 4{x^2}{y^2} + 4{x^3}y\\
= \left( {7{x^2}{y^2} - 3{x^2}{y^2} - 4{x^2}{y^2}} \right) + \left( { - 7{x^3}y + 4{x^3}y} \right) + 3{y^3}\\
= - 3{x^3}y + 3{y^3}\\
x = 1;y = \dfrac{1}{2} \Rightarrow B = - {3.1^3}.\dfrac{1}{2} + 3.{\left( {\dfrac{1}{2}} \right)^3} = - \dfrac{3}{2} + 3.\dfrac{1}{8} = - \dfrac{9}{8}
\end{array}\)
