Question

Cho A=3x+2/x-3

a) Tìm x nguyên để A là sốhữu tỉ

b) Tìm x nguyên để A là số nguyên

Answer 1

\(b,A=\frac{3x+2}{x-3}\)\(=\frac{x-3+2x-6+11}{x-3}\)\(=\frac{\left(x-3\right)+2\left(x-3\right)+11}{x-3}\)\(=\frac{x-3}{x-3}+\frac{2\left(x-3\right)}{x-3}+\frac{11}{x-3}\)\(=1+2+\frac{11}{x-3}\)\(=3+\frac{11}{x-3}\)

Để A nguyên => \(\frac{11}{x-3}\)nguyên => \(11⋮x-3\)\(\Rightarrow x-3\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

Ta có bảng sau:

x-3	-11	-1	1	11
x	-8	2	4	14

Vậy................