1.
\(A< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow0\le x< 4\)
Bài 2:
Để \(A\ge1\) thì \(A-1\ge0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\sqrt{x}-1>0\)
hay x>1
2.
\(A\ge1\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\ge1\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}-1\ge0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\sqrt{x}-1>0\)
\(\Leftrightarrow x>1\)
