Đáp án:
\(\begin{array}{l}
1,\\
a,\\
3{x^3} + 3{x^2}y - 24x\\
b,\\
2{x^2} - 9{y^2}\\
c,\\
2x{y^2} + \dfrac{5}{2}{x^2}{y^2} - 3xy\\
d,\\
- 2{x^2}y - 3{x^2}{y^2} + 8x{y^2}\\
2,
\end{array}\)
\(\begin{array}{l}
a,\\
A = - 3{x^3} + 6{x^2} + 6\\
x = 5 \Rightarrow A = - 219\\
b,\\
B = - 3{x^3}y + 3{y^3}\\
x = 1;y = \dfrac{1}{2} \Rightarrow B = - \dfrac{9}{8}
\end{array}\)
\(\begin{array}{l}
3,\\
a,\\
x = 12\\
b,\\
x = 4
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
3x\left( {{x^2} + xy - 8} \right)\\
= 3x.{x^2} + 3x.xy - 3x.8\\
= 3{x^3} + 3{x^2}y - 24x\\
b,\\
3y\left( {2x - 3y} \right) + 2x\left( {x - 3y} \right)\\
= 3y.2x - 3y.3y + 2x.x - 2x.3y\\
= 6xy - 9{y^2} + 2{x^2} - 6xy\\
= 2{x^2} - 9{y^2}\\
c,\\
\left( {4y + 5xy - 6} \right).\dfrac{1}{2}xy\\
= 4y.\dfrac{1}{2}xy + 5xy.\dfrac{1}{2}xy - 6.\dfrac{1}{2}xy\\
= 2x{y^2} + \dfrac{5}{2}{x^2}{y^2} - 3xy\\
d,\\
- xy\left( {2x + 3xy - 8y} \right)\\
= - xy.2x - xy.3xy - xy.\left( { - 8y} \right)\\
= - 2{x^2}y - 3{x^2}{y^2} + 8x{y^2}\\
2,
\end{array}\)
\(\begin{array}{l}
a,\\
A = 3\left( {3{x^2} + 2} \right) - {x^2}.\left( {4x + 3} \right) + {x^3}\\
= \left( {9{x^2} + 6} \right) - \left( {4{x^3} + 3{x^2}} \right) + {x^3}\\
= 9{x^2} + 6 - 4{x^3} - 3{x^2} + {x^3}\\
= \left( { - 4{x^3} + {x^3}} \right) + \left( {9{x^2} - 3{x^2}} \right) + 6\\
= - 3{x^3} + 6{x^2} + 6\\
x = 5 \Rightarrow A = - {3.5^3} + {6.5^2} + 6 = - 3.125 + 6.25 + 6 = - 375 + 150 + 6 = - 219\\
b,\\
B = 7xy\left( {xy - {x^2}} \right) - 3{y^2}\left( {{x^2} - y} \right) - 4{x^2}\left( {{y^2} - xy} \right)\\
= \left( {7{x^2}{y^2} - 7{x^3}y} \right) - \left( {3{x^2}{y^2} - 3{y^3}} \right) - \left( {4{x^2}{y^2} - 4{x^3}y} \right)\\
= 7{x^2}{y^2} - 7{x^3}y - 3{x^2}{y^2} + 3{y^3} - 4{x^2}{y^2} + 4{x^3}y\\
= \left( {7{x^2}{y^2} - 3{x^2}{y^2} - 4{x^2}{y^2}} \right) + \left( { - 7{x^3}y + 4{x^3}y} \right) + 3{y^3}\\
= - 3{x^3}y + 3{y^3}\\
x = 1;y = \dfrac{1}{2} \Rightarrow B = - {3.1^3}.\dfrac{1}{2} + 3.{\left( {\dfrac{1}{2}} \right)^3} = - \dfrac{3}{2} + 3.\dfrac{1}{8} = - \dfrac{9}{8}
\end{array}\)
\(\begin{array}{l}
3,\\
a,\\
3x\left( {x + 1} \right) - x\left( {3x + 2} \right) = 12\\
\Leftrightarrow \left( {3{x^2} + 3x} \right) - \left( {3{x^2} + 2x} \right) = 12\\
\Leftrightarrow 3{x^2} + 3x - 3{x^2} - 2x = 12\\
\Leftrightarrow \left( {3{x^2} - 3{x^2}} \right) + \left( {3x - 2x} \right) = 12\\
\Leftrightarrow x = 12\\
b,\\
{x^2}\left( {4 - 3x} \right) + x\left( {3{x^2} - 4x + 2} \right) = 8\\
\Leftrightarrow \left( {4{x^2} - 3{x^3}} \right) + \left( {3{x^3} - 4{x^2} + 2x} \right) = 8\\
\Leftrightarrow 4{x^2} - 3{x^3} + 3{x^3} - 4{x^2} + 2x = 8\\
\Leftrightarrow \left( { - 3{x^3} + 3{x^3}} \right) + \left( {4{x^2} - 4{x^2}} \right) + 2x = 8\\
\Leftrightarrow 2x = 8\\
\Leftrightarrow x = 4
\end{array}\)
