Question

tìm x nguyên để A đặt giá trị nguyên

A = x + 2 / x^2 - x +3

Answer 1

\(A=\dfrac{x+2}{x^2-x+3}\Leftrightarrow Ax^2-Ax+3A=x+2\\ \Leftrightarrow Ax^2-x\left(A+1\right)+3A-2=0\\ \Leftrightarrow\Delta=\left(A+1\right)^2-4A\left(3A-2\right)\ge0\\ \Leftrightarrow-11A+10A+1\ge0\\ \Leftrightarrow-\dfrac{1}{11}\le A\le1\)

Mà \(A\in Z\Leftrightarrow A\in\left\{0;1\right\}\)

\(+)A=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\\ +)A=1\Leftrightarrow x+2=x^2-x+3\Leftrightarrow x=1\)

Vậy \(x\in\left\{-2;1\right\}\Leftrightarrow A\in Z\)