a: Thay x=9 vào P, ta được:
\(P=\dfrac{9+3}{\sqrt{9}-2}=\dfrac{12}{3-2}=12\)
b: \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c: Đặt A=P:Q
\(=\dfrac{x+3}{\sqrt{x}-2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{3}{\sqrt{x}}}=2\sqrt{3}\)
Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=3\)
=>x=3(nhận)
