Question

\(\dfrac{\sqrt{x}-1}{2\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-3}< \dfrac{1}{2}\)

trình bày bước

Answer 1

\(\Rightarrow\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}-3\right)}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{\sqrt{x}-1}{2\sqrt{x}-6}< \dfrac{1}{2}\)

\(\Rightarrow\dfrac{\sqrt{x}-1}{2\sqrt{x}-6}-\dfrac{1}{2}< 0\)

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

\(\Rightarrow\dfrac{2}{2\left(\sqrt{x}-3\right)}< 0\)

\(\Rightarrow\sqrt{x}-3< 0\)

\(\Rightarrow\sqrt{x}< 3\)

=> 0 < x < 9

Answer 2

\(ĐK:x>0;x\ne9\\ BPT\Leftrightarrow\dfrac{\sqrt{x}-1}{2}\cdot\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{2}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}-3\right)}-\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\\ \Leftrightarrow\dfrac{2}{2\left(\sqrt{x}-3\right)}< 0\Leftrightarrow\sqrt{x}-3< 0\left(2>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0< x< 9\left(ĐKXĐ\right)\)