Question

Tìm \(x\in Z\)

để A=\(\frac{1-2x}{x+3}\in Z\)

Answer 1

Ta có:

\(A=\frac{1-2x}{x+3}=\frac{1-2x-6+6}{x+3}=\frac{1-\left(2x+6\right)+6}{x+3}=\frac{\left(1+6\right)-\left(2x+6\right)}{x+3}=\frac{7-\left(2x+2\times3\right)}{x+3}=\frac{7-2\times\left(x+3\right)}{x+3}=\frac{7}{x+3}-\frac{2\times\left(x+3\right)}{x+3}=\frac{7}{x+3}-2\)

Để \(A=\frac{1-2x}{x+3}\in Z\) \(\left(x\ne-3\right)\)

thì \(\frac{7}{x+3}\in Z\) \(\left(x\ne-3\right)\)

\(\Rightarrow x+3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Ta có bảng sau:

\(x+3\)	\(-7\)	\(-1\)	\(1\)	\(7\)
\(x\)	\(-10\)	\(-4\)	\(-2\)	\(4\)

mà \(x\in Z\) và \(x\ne-3\)

\(\Rightarrow x\in\left\{-10;-4;-2;4\right\}\)

Vậy \(x\in\left\{-10;-4;-2;4\right\}\) thì thỏa mãn đề bài.