Question

cho P=\(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\).Tìm x hữu tỉ để \(\dfrac{9}{P}\) là số nguyên

Answer 1

Ta có:\(P=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\left(DKXĐ:x\ge0\right)\)

Ta có:\(\left(\sqrt{x}-2\right)^2\ge0\)

\(\Leftrightarrow x+4\ge4\sqrt{x}\)

\(\Leftrightarrow x+4\sqrt{x}+4\ge8\sqrt{x}\Leftrightarrow\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\ge8\)

\(\Rightarrow0< \dfrac{9}{P}\le\dfrac{9}{8}\)

Để \(\dfrac{9}{P}\in Z\) thì \(\dfrac{9}{P}=1\Leftrightarrow P=9\)

Với P=9\(\Leftrightarrow\left(\sqrt{x}+2\right)^2=9\sqrt{x}\)

\(\Leftrightarrow x-5\sqrt{x}+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\)