Question

Bài 1: Tìm x nguyên để biểu thức A= \(\dfrac{1}{x-1}:\dfrac{x-2}{2\left(x-1\right)}\) có giá trị nguyên

Answer 1

\(A=\dfrac{1}{x-1}:\dfrac{x-2}{2\left(x-1\right)}=\dfrac{1}{x-1}.\dfrac{2\left(x-1\right)}{x-2}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\dfrac{2}{x-2}\)Để \(A\in Z\) thì \(2⋮x-2\Leftrightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=4\\x=0\end{matrix}\right.\left(TM\right)\)