Question

cho A=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\dfrac{x^2-2x+1}{2}\)

a) rút gọn A

b) tìm x để A>0

c) tìm gtln của a

Answer 1

a: \(A=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2}{2}\)

\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b: Để A>0 thì \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\)

=>căn x-1<0

=>0<x<1

c: \(A=-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)

Dấu = xảy ra khi x=1/4