Ta có :
\(\sqrt{x}+1=\sqrt{x}-3+4\)chia hết cho \(\sqrt{x}-3\)\(\Rightarrow\)\(4\)chia hết cho \(\sqrt{x}-3\)\(\Rightarrow\)\(\left(\sqrt{x}-3\right)\inƯ\left(4\right)\)
Mà \(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Do đó :
\(x\in\left\{16;4;25;1;49\right\}\)
Vậy \(x\in\left\{16;4;25;1;49\right\}\)thì biểu thức \(A\)là số nguyên
\(A=\frac{\sqrt{x+1}}{\sqrt{x+3}}=\sqrt{\frac{x+1}{x-3}}=\sqrt{\frac{x-1}{x-3}}+\sqrt{\frac{4}{x-3}}=1+\frac{2}{\sqrt{x-3}}\)
Để A nguyên thì: \(\sqrt{x-3}\inƯ\left(2\right)\)
Mà \(Ư\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) \(\sqrt{x-3}=^+_-1\Rightarrow x-3=1\Rightarrow x=4\)
+) \(\sqrt{x-3}=^+_-2\Rightarrow x-3=4\Rightarrow x=7\)
Vậy: x = {4; 7} thì A nguyên
