Đáp án:
\(\left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A\left( x \right) = 7{x^2} - 8{x^3} + 3{x^2} - \dfrac{1}{2}x + 21 + {x^3}\\
= - 7{x^3} + 10{x^2} - \dfrac{1}{2}x + 21\\
B\left( x \right) = - 8{x^2} + 3{x^3} - 2{x^2} + \dfrac{1}{2}x + 2{x^3} - 5\\
= 5{x^3} - 10{x^2} + \dfrac{1}{2}x - 5\\
b)H\left( x \right) = A\left( x \right) + B\left( x \right)\\
= - 7{x^3} + 10{x^2} - \dfrac{1}{2}x + 21 + 5{x^3} - 10{x^2} + \dfrac{1}{2}x - 5\\
= - 2{x^3} + 16\\
K\left( x \right) = \left( {x + 3} \right).H\left( x \right) = 0\\
\to \left( {x + 3} \right)\left( { - 2{x^3} + 16} \right) = 0\\
\to - 2\left( {x + 3} \right)\left( {{x^3} - 8} \right) = 0\\
\to \left( {x + 3} \right)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) = 0\\
\to \left( {x + 3} \right)\left( {x - 2} \right) = 0\left( {do:{x^2} + 2x + 4 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.
\end{array}\)
