a) Thay x=9 vào B ta có:
\(B=\dfrac{4\left(\sqrt{9}+2\right)}{\sqrt{9}-2}=\dfrac{4\cdot\left(3+2\right)}{3-2}=\dfrac{4\cdot5}{1}=20\)
b) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}+\dfrac{4x}{x-4}\)
\(A=\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{8\sqrt{x}+4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)
c) \(P=\dfrac{A}{B}\)
\(P=\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}:\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)
\(P=\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{4\sqrt{x}}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
Với \(x>0\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}}>2\)
\(\Rightarrow P\ge2\) thì \(P>\sqrt{P}\)
a: Khi x=9 thì \(B=\dfrac{4\left(3+2\right)}{3-2}=4\cdot5=20\)
b: \(A=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{x-4}=\dfrac{4x+8\sqrt{x}}{x-4}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)
c: P=A/B
\(=\dfrac{4\sqrt{x}}{\sqrt{x}-2}:\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
căn x<căn x+2
=>P=căn x/căn x+2<1
căn x>=0; căn x+2>0
=>P>=0
=>0<=P<1
=>P<căn P
