Question

Tìm `m, n` để `A(x)=(m+3)x^2-(2n-1)x-1` đồng thời chia hết cho `x+1 ;x-3`

Answer 1

\(A\left(x\right)\) đồng thời chia hết \(x+1;x-3\)

\(\Rightarrow A\left(x\right)\) nhận \(x=-1;x=3\) là 2 nghiệm

Thay vào ta được: \(\left\{{}\begin{matrix}\left(m+3\right).\left(-1\right)^2-\left(2n-1\right).\left(-1\right)-1=0\\\left(m+3\right).3^2-\left(2n-1\right).3-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m+2n=-1\\9m-6n=-29\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3m+6n=-3\\9m-6n=-29\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}12m=-32\\9m-6n=-29\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m=-\dfrac{8}{3}\\n=\dfrac{5}{6}\end{matrix}\right.\)