A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+3+1}{\sqrt{x}-3}+\dfrac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)
A nguyên => \(\sqrt{x}-3=U\left(4\right)=\left\{-4,-2,-1,1,2,4\right\}\)
\(\sqrt{x}=\left\{1,4\right\}\Rightarrow x=\left\{1,2\right\}\)
Phạm Đức Minh m gửi câu hỏi ở tận đề 16 luôn ak