Answer 1

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{3}{4}\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3}{4}< 0\)

\(\Leftrightarrow\dfrac{4\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-4-3\sqrt{x}-3}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-7}{4\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-7< 0\)

\(\Leftrightarrow x< 49\)

Và điều kiện nên ta kết luận: \(0\le x< 49\)

Answer 2

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3}{4}< 0\)

=>\(\dfrac{4\sqrt{x}-4-3\sqrt{x}-3}{4\left(\sqrt{x}+1\right)}< 0\)

=>căn x-7<0

=>0<x<49