Question

em cần gấp bài 2 ạ, em cảm ơn

Answer 1

Bài 2:

Ta có:

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\) 

A nhận giá trị nguyên khi \(\dfrac{4}{\sqrt{x}-3}\) nguyên:

\(\Rightarrow4\) ⋮ \(\sqrt{x}-3\) 

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(\sqrt{x}-3\ge-3\)

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

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

\(\Rightarrow x\in\left\{16;4;25;1;49\right\}\)