Question

giúp mik với mn

Answer 1

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

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

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

Để \(A< \dfrac{2}{3}\) thì \(A-\dfrac{2}{3}< 0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2}{3}< 0\)

=>\(\dfrac{3\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+2\right)}{3\left(\sqrt{x}+2\right)}< 0\)

=>\(3\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+2\right)< 0\)

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

=>\(\sqrt{x}< 1\)

=>0<=x<1

Kết hợp ĐKXĐ, ta được: 0<x<1