Đáp án:
\(b.B\left( x \right) = - 3{x^3} + \dfrac{{15}}{2}{x^2} + \dfrac{7}{2}x - \dfrac{{25}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.M\left( x \right) = \dfrac{5}{2}{x^2} - \dfrac{1}{2}x - {x^3} - 1\\
N\left( x \right) = - 2{x^3} + 5{x^2} - 12 + 4x\\
\to A\left( x \right) = M\left( x \right) - N\left( x \right)\\
= \dfrac{5}{2}{x^2} - \dfrac{1}{2}x - {x^3} - 1 + 2{x^3} - 5{x^2} + 12 - 4x\\
= {x^3} - \dfrac{5}{2}{x^2} - \dfrac{9}{2}x + 11\\
A\left( x \right) = 0\\
\to {x^3} - \dfrac{5}{2}{x^2} - \dfrac{9}{2}x + 11 = 0\\
\to 2{x^3} - 5{x^2} - 9x + 22 = 0\\
\to 2{x^3} - 4{x^2} - {x^2} + 2x - 11x + 22 = 0\\
\to 2{x^2}\left( {x - 2} \right) - x\left( {x - 2} \right) - 11\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
2{x^2} - x - 11 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{{1 + \sqrt {89} }}{4}\\
x = \dfrac{{1 - \sqrt {89} }}{4}
\end{array} \right.\\
b.B\left( x \right) = M\left( x \right) + N\left( x \right)\\
= \dfrac{5}{2}{x^2} - \dfrac{1}{2}x - {x^3} - 1 - 2{x^3} + 5{x^2} - 12 + 4x\\
= - 3{x^3} + \dfrac{{15}}{2}{x^2} + \dfrac{7}{2}x - \dfrac{{25}}{2}
\end{array}\)
⇒ Bậc 3
