Question

Tìm x thuộc Z để A thuộc Z và tìm giá trị đó

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

Answer 1

A = \(\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\)

Vì A \(\in\) Z => \(\dfrac{5}{x-2}\in\) Z

hay 5 chia hết cho x-2

=> x-2 \(\in\)Ư₍₅₎

Ta có bảng

x-2 -5 -1 1 5

x -3(TM) 1(TM) 3(TM) 7(TM)

Vậy x \(\in\){ -3;1;3;7}

Answer 2

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

=> \(A=\dfrac{x-2+5}{x-2}=1+\dfrac{5}{x-2}\)

=> 5\(⋮x-2\)

Hay \(x-2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

=> \(\left[{}\begin{matrix}x-2=-1=>x=1\\x-2=1=>x=3\\x-2=-5=>x=-3\\x-2=5=>x=7\end{matrix}\right.\)

Mà \(x\in z\)

=> dpcm