Question

cho B = \(\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right)\) : \(\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
mình rút gọn được \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
 Chứng minh B > \(\dfrac{1}{3}\)

Answer 1

a: \(B=\dfrac{x+3+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

b: \(B-\dfrac{1}{3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{3}=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{\sqrt{x}+3}>0\)

=>B>1/3