Question

Cho A=\(\dfrac{2\sqrt{x}+1}{\sqrt{x}}\) và B=\(\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\) với x>0,x≠4

a.rút gọn B

B.Cho P=B/A.Tìm x để P<\(|P|\)

Answer 1

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

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

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

 b: P=B:A

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

Để P<|P| thì P<0

=>\(\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

kết hợp ĐKXĐ, ta được: 0<x<4