Question

Answer 1

1: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3}=\dfrac{5}{3}\)

2: \(B=\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-3\sqrt{x}+4}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\)

\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

 

3: \(P=A:B=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2+4}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)

Để P max thì \(\sqrt{x}-2=1\)

=>x=9