Question

Tìm các số nguyên x để số hữu tỉ sau là số nguyên

\(t=\dfrac{x+7}{x-3}\) \(t=\dfrac{2x+7}{x+3}\)

Answer 1

\(\dfrac{x+7}{x-3}\)là số nguyên

⇒\(\left(x+7\right)\) ⋮ \(\left(x-3\right)\)

\(\left(x-3\right)\) ⋮ \(\left(x-3\right)\)

⇒\(\left(x+7\right)-\left(x-3\right)\)⋮\(\left(x-3\right)\)

⇒10 ⋮ \(\left(x-3\right)\)

⇒\(\left(x-3\right)\) ∈ \(Ư\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

⇒\(x\) ∈ \(\left\{4;2;5;1;8;-2;13;-7\right\}\)

Answer 2

\(\dfrac{2x+7}{x+3}\) là số nguyên

⇒ \(2x+7\) ⋮ \(x+3\)

\(x+3\) ⋮ \(x+3\) ⇒ \(2.\left(x+3\right)\)= \(2x+6\)

⇒ \(\left(2x+7\right)-\left(2x+6\right)\)⋮ \(\left(x+3\right)\)

⇒ \(1\) ⋮ \(\left(x+3\right)\)

⇒\(\left(x+3\right)\) ∈ \(Ư\left(1\right)=\left\{1;-1\right\}\)

⇒ \(x\) ∈ \(\left\{-2;-4\right\}\)