Question

Cho B = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

Tìm x \(\in\) Z để B có giá trị nguyên .

Answer 1

Ta có : \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để \(B\in Z\Leftrightarrow\frac{4}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow4⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1,\pm2,\pm4\right\}\)

Do : \(\sqrt{x}\ge0\Rightarrow\sqrt{x}-3\ge-3\)

\(\Rightarrow\sqrt{x}-3\in\left\{1,-1,2,-2,4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4,2,5,1,7\right\}\)

\(\Rightarrow x\in\left\{2,\sqrt{2},\sqrt{5},\sqrt{1},\sqrt{7}\right\}\)

mà : \(x\in Z\Rightarrow x=2\)

Vậy : \(x=2\) thì \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}\) có giá trị nguyên.