Question

\(N=\frac{\sqrt{x}+1}{\sqrt{x-3}}\)TÌM X\(\in\)Z ĐỂ N LÀ SỐ NGUYÊN DƯƠNG

Answer 1

x-3=t^2

​N dương=>t>0

​N=(t^2+3)/t=t+3/t

​t={,1 ,3)

​=>x={4}

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​N=(|k|+1|/(|k|-1

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Answer 2

\(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}+3}\)

Để N thuộc N

\(\Rightarrow4⋮\sqrt{x}-3\)

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

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

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

\(\Rightarrow x\in\left(2;-2;1;-1\right)\)