ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{-\left(\sqrt{x}-1\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
\(P=\frac{1}{2}\Rightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\Rightarrow4-10\sqrt{x}=\sqrt{x}+3\)
\(\Rightarrow11\sqrt{x}=1\Rightarrow\sqrt{x}=\frac{1}{11}\Rightarrow x=\frac{1}{121}\)
\(P=\frac{17-5\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=-5+\frac{17}{\sqrt{x}+3}\)
Do \(\sqrt{x}+3\ge3\Rightarrow-5+\frac{17}{\sqrt{x}+3}\le-5+\frac{17}{3}=\frac{2}{3}\)
\(P_{max}=\frac{2}{3}\) khi \(x=0\), hình như bạn nhầm đề, ko có GTNN đâu, chỉ có GTLN thôi
\(P=-5+\frac{17}{\sqrt{x}+3}\Rightarrow\) để P nguyên thì \(\sqrt{x}+3=Ư\left(17\right)\)
Mà \(\sqrt{x}+3\ge3\Rightarrow\sqrt{x}+3=17\)
\(\Rightarrow\sqrt{x}=14\Rightarrow x=196\)
