1. Em tự giải
2.
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
3.
\(P=A.B=\dfrac{7}{\sqrt{x}+3}\)
Ta có: \(\left\{{}\begin{matrix}7>0\\\sqrt{x}+3>0\end{matrix}\right.\) \(\Rightarrow P>0\)
Và \(\sqrt{x}+3\ge3\Rightarrow\dfrac{7}{\sqrt{x}+3}\le\dfrac{7}{3}\)
\(\Rightarrow0< P\le\dfrac{7}{3}\)
Mà P nguyên \(\Rightarrow\left[{}\begin{matrix}P=1\\P=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{\sqrt{x}+3}=1\\\dfrac{7}{\sqrt{x}+3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=7\\2\sqrt{x}+6=7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=16\\x=\dfrac{1}{4}\end{matrix}\right.\)
