a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Vì x=9 thỏa mãn ĐKXĐ nên thay x=9 vào biểu thức \(P=\frac{x+3}{\sqrt{x}-2}\), ta được:
\(P=\frac{9+3}{\sqrt{9}-2}=\frac{12}{3-2}=\frac{12}{1}=12\)
Vậy: Khi x=9 thì P=12
b) Ta có: \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c) Ta có: \(\frac{P}{Q}=\frac{x+3}{\sqrt{x}-2}:\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{x+3}{\sqrt{x}}=\sqrt{x}+\frac{3}{\sqrt{x}}\ge2\cdot\sqrt{\sqrt{x}\cdot\frac{3}{\sqrt{x}}}=2\sqrt{3}\)
\(\Leftrightarrow\frac{P}{Q}\ge2\sqrt{3}\)
Dấu '=' xảy ra khi \(\frac{x+3}{\sqrt{x}}=2\sqrt{3}\)
\(\Leftrightarrow x+3=2\sqrt{3}\cdot\sqrt{x}\)
\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot\sqrt{3}+3=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3}\)
hay x=3(nhận)
Vậy: Khi x=3 thì biểu thức \(\frac{P}{Q}\) đạt giá trị nhỏ nhất
