Question

Xác định số hữu tỉ a,b sao cho:

a) f(x)=\(^{2x^3}\)- \(_{x^2}\)+ax +b chia hết cho \(^{x^2}\)- 1

b) f(x)=\(^{3x^3}\) + \(^{\text{ax}^2}\)+ bx +9 chia hết cho \(^{x^2}\)- 9

Answer 1

a/ \(f\left(x\right)⋮\left(x^2-1\right)\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=0\\f\left(-1\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2-1+a+b=0\\-2-1-a+b=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\-a+b=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-2\\b=1\end{matrix}\right.\)

b/ Tương tự câu a, ta có \(\left\{{}\begin{matrix}f\left(3\right)=0\\f\left(-3\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}9a+3b=-90\\9a-3b=72\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-27\end{matrix}\right.\)