Đáp án:
b) \(\left[ \begin{array}{l}
x = 0\\
x = \dfrac{9}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)f\left( x \right) = \left( { - 3 + 3} \right){x^4} + 4{x^3} + \left( { - 7 + 2} \right){x^2} + 3x + \dfrac{1}{2} - \dfrac{9}{2}\\
= 4{x^3} - 5{x^2} + 3x - 4\\
g\left( x \right) = 4{x^3} - 3{x^2} - 6x - 4\\
b)h\left( x \right) + g\left( x \right) = f\left( x \right)\\
\to h\left( x \right) = f\left( x \right) - g\left( x \right)\\
= 4{x^3} - 5{x^2} + 3x - 4 - 4{x^3} + 3{x^2} + 6x + 4\\
= - 2{x^2} + 9x\\
h\left( x \right) = 0 \to - 2{x^2} + 9x = 0\\
\to x\left( { - 2x + 9} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{9}{2}
\end{array} \right.
\end{array}\)
