ĐKXĐ của P là x>0; x<>9
\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{7\sqrt{x}+3}{x-9}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\sqrt{x}\left(\sqrt{x}-3\right)-7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+4\sqrt{x}+3+2x-6\sqrt{x}-7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)
\(P=A\cdot B=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{x+7}{3\sqrt{x}}=\dfrac{x+7}{\sqrt{x}+3}\)
=>\(P=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)
=>\(P=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{16}{\sqrt{x}+3}}-6=2\cdot4-6=2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}+3=\sqrt{16}=4\)
=>x=1(nhận)
