Question

Tìm x nguyên để \(\frac{2\sqrt{x}-3}{\sqrt{x}+1}\) là số nguyên

Answer 1

đkxđ : x\(\ge\)\(\)0

Để \(\frac{2\sqrt{x}-3}{\sqrt{x}+1}\) là số nguyên thì \(\left(2\sqrt{x}-3\right)⋮\left(\sqrt{x}+1\right)\)

Ta có \(\frac{2\sqrt{x}-3}{\sqrt{x}+1}=2+\frac{-5}{\sqrt{x}+1}\Rightarrow\left(-5\right)⋮\left(\sqrt{x}+1\right)\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(-5\right)\)

Mà Ư(-5)=\(\left\{1,-1,5,-5\right\}\)

Do \(\sqrt{x}+1\ge1\Rightarrow\left(\sqrt{x}+1\right)\in\left\{1,5\right\}\Rightarrow x\in\left\{0,16\right\}\)