Question

tìm x nguyen de A = \(\frac{\sqrt{x}}{\sqrt{x}-1}\)ngyen

 

Answer 1

Ta có\(\frac{\sqrt{x}}{\sqrt{x}-1}=\frac{\sqrt{x}-1+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-1}=1+\frac{1}{\sqrt{x}-1}\)

Vậy để A nguyên thì \(1+\frac{1}{\sqrt{x}-1}\)nguyên

hay \(\frac{1}{\sqrt{x}-1}\)nguyên

hay \(1⋮\sqrt{x}-1\)

hay \(\sqrt{x}-1\in\left\{1;-1\right\}\)

*Với \(\sqrt{x}-1=1\)

thì \(\sqrt{x}=2\)

Hay \(x=4\)

*Với \(\sqrt{x}-1=-1\)

thì \(\sqrt{x}=0\)

hay \(x=0\)

Answer 2

để \(\frac{\sqrt{x}}{\sqrt{x}-1}\)là số nguyên thì  \(\sqrt{x}-1=1\)

=> \(\sqrt{x}=2\)

=>\(x=2^2\)

=> \(x=4\)