Để \(f\left(x\right)=x^2-\left(m+1\right)x+4\) chia hết cho \(x-1\) thì: \(f\left(1\right)=1^2-\left(m+1\right)+4=0\)
\(\Leftrightarrow1-m-1+4=0\)
\(\Leftrightarrow4-m=0\)
\(\Rightarrow m=4\)
Vậy: với m = 4 thì \(x^2-\left(m+1\right)x+4⋮\left(x-1\right)\)
\(b,g\left(x\right)=x^2-3x+2=x^2-2x-x+2\)
\(=x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
Để \(f\left(x\right)\) đồng thời chia hết cho \(\left(x-1\right);\left(x-2\right)\) thì:
\(\left\{{}\begin{matrix}f\left(1\right)=1^4-5+a=0\\f\left(2\right)=2^4-5.2^2+a=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f\left(1\right)=a-4=0\\f\left(2\right)=a-4=0\end{matrix}\right.\)
\(\Rightarrow a=4\)
Vậy :....
Cái này mk chắc 50% thôi . Sai thôg cảm nhá>.<
