Question

Tim x để \(P=\dfrac{9}{\sqrt{x}+4}\)có giá trị là số nguyên

 

Answer 1

Do \(\left\{{}\begin{matrix}9>0\\\sqrt{x}+4>0\end{matrix}\right.\) \(\Rightarrow\dfrac{9}{\sqrt{x}+4}>0\)

\(\left\{{}\begin{matrix}9>0\\\sqrt{x}+4\ge4;\forall x\ge0\end{matrix}\right.\) \(\Rightarrow\dfrac{9}{\sqrt{x}+4}\le\dfrac{9}{4}\)

\(\Rightarrow0< P\le\dfrac{9}{4}\)

Mà P nguyên \(\Rightarrow\left[{}\begin{matrix}P=1\\P=2\end{matrix}\right.\)

Với \(P=1\Rightarrow\dfrac{9}{\sqrt{x}+4}=1\Rightarrow\sqrt{x}=5\Rightarrow x=25\)

Với \(P=2\Rightarrow\dfrac{9}{\sqrt{x}+4}=2\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

Vậy \(x=\left\{\dfrac{1}{4};25\right\}\) thì P nguyên