Question

A= \(\dfrac{x}{x-3}\)
a) tính A biết lx-2l=1
b) tìm x∈Z để A∈Z

Answer 1

\(ĐK:x\ne3\\ a,\left|x-2\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1+2=3\left(ktm\right)\\x=-1+2=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\\ \Leftrightarrow A=\dfrac{1}{1-3}=-\dfrac{1}{2}\\ b,A=\dfrac{x-3+3}{x-3}=1+\dfrac{3}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{0;2;4;6\right\}\)

Answer 2

ĐKXĐ: \(x\ne3\)

a) \(\left|x-2\right|=1\)\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

\(A=\dfrac{x}{x-3}=\dfrac{1}{1-3}=-\dfrac{1}{2}\)

b) \(A=\dfrac{x-3+3}{x-3}=1+\dfrac{3}{x-3}\in Z\)

\(\Rightarrow\left(x-3\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x\in\left\{0;2;4;6\right\}\)