mình đang cần gấppppp
\(A\ge1\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}-1\ge0\\ \Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\ge0\\ \Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
để
\(A\ge1\Rightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\ge1\)
nếu \(0\le x< 1\) thì \(\sqrt{x}-1< 0\\ \Rightarrow2\sqrt{x}-1< \sqrt{x}-1\\ \Rightarrow\sqrt{x}< 0\)
vô lý,loại
nếu x>1 thì \(\sqrt{x}-1>0\)
\(\Rightarrow2\sqrt{x}-1>\sqrt{x}-1\\ \Rightarrow\sqrt{x}>0\Rightarrow x>1\)
vậy để A>/=1 thì x>1
Để \(A\ge1\) thì \(A-1\ge0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}-1\ge0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}+1}{\sqrt{x}-1}\ge0\)
\(\Leftrightarrow\sqrt{x}-1>0\)
hay x>1
