Question

Tìm giá trị nguyên của x để Px ⋮ Qx biết

Px = x³ - x² +2

Qx = x - 1

Answer 1

\(\dfrac{P_x}{Q_x}=\dfrac{x^3-x^2+2}{x-1}=\dfrac{x^3-x^2}{x-1}+\dfrac{2}{x-1}\)

\(\dfrac{P_x}{Q_x}=\dfrac{x^2\left(x-1\right)}{x-1}+\dfrac{2}{x-1}=x^2+\dfrac{2}{x-1}\)

Để \(P_x⋮Q_x\) thì \(\left(x-1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\\x=-1\end{matrix}\right.\)