Question

Toán nâng cao !!

Answer 1

\(A=\dfrac{x+15}{x-2}=\dfrac{x-2+17}{x-2}\)     \(\left(ĐK:x\ne2\right)\)

\(=\dfrac{x-2}{x-2}+\dfrac{17}{x-2}=1+\dfrac{17}{x-2}\)

Để \(A=1+\dfrac{17}{x-2}\in Z\)

thì \(\dfrac{17}{x-2}\in Z\)

\(\Rightarrow17⋮x-2\)

\(\Rightarrow x-2\inƯ\left(17\right)\)

\(\Rightarrow x-2\in\left\{1;17;-1;-17\right\}\)

\(\Rightarrow x\in\left\{3;19;1;-15\right\}\left(tm\right)\)

\(Vậy:x\in\left\{3;19;1;-15\right\}\)

Answer 2

Ta có:

x + 15 = x - 2 + 17

Để A ∈ Z thì (x + 15) ⋮ (x - 2)

⇒ 17 ⋮ (x - 2)

⇒ x - 2 ∈ Ư(17) = {-17; -1; 1; 17}

⇒ x ∈ {-15; 1; 3; 19}