Question

Tìm XϵR để \(B=\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\)  thuộc Z

Answer 1

\(B=\dfrac{\sqrt{x}-1}{3-\sqrt{x}}\);\(x\ge0;x\ne9\)

\(B=-\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)

\(B=-\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}\)

\(B=-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}-\dfrac{2}{\sqrt{x}-3}\)

\(B=-1-\dfrac{2}{\sqrt{x}-3}\)

Để B nguyên thì \(\dfrac{2}{\sqrt{x}-3}\in Z\) hay \(\sqrt{x}-3\in U\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(*\)\(\sqrt{x}-3=1\rightarrow x=16\)

\(*\)\(\sqrt{x}-3=-1\rightarrow x=4\) 

\(*\)\(\sqrt{x}-3=2\rightarrow x=25\)

\(*\)\(\sqrt{x}-3=-2\rightarrow x=1\)

Vậy \(x=\left\{16;4;25;1\right\}\) thì B là số nguyên