Question

Tìm x thuộc Z để A thuộc Z. 

A= \(\frac{1-2x}{x+3}\)

 

Answer 1

Ta thấy:
\(A=\frac{1-2x}{x+3}=\frac{7-6-2x}{x+3}=\frac{7-\left(6+2x\right)}{x+3}=\frac{7-2\left(3+x\right)}{x+3}=\frac{7}{x+3}-\frac{2\left(3+x\right)}{x+3}=\frac{7}{x+3}-2\)
Do \(2\in Z\), để \(A=\frac{1-2x}{x+3}\in Z\) thì \(\frac{7}{x+3}\in Z\)
\(\Rightarrow x+3\in U\left(7\right)=\left\{-7;-1;1;7\right\}\)
* TH1: x + 3 = -7    =>    x = -10
* TH2: x + 3 = -1    =>    x = -4
* TH3: x + 3 = 1     =>    x = -2
* TH4: x + 3 = 7     =>    x = 4
Vậy \(x\in\left\{-10;-4;-2;4\right\}\)