1) \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\)( đk x>0)
\(\Leftrightarrow\dfrac{2\sqrt{x}-4-\sqrt{x}}{6\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-4}{6\sqrt{x}}>0\)
vì \(6\sqrt{x}>0\Rightarrow\sqrt{x}-4>0\)
\(\Leftrightarrow x>16\)
2) theo cauchy schwars:
\(x+1\ge2\sqrt{x}\)
\(\Leftrightarrow x+\sqrt{x}+1\ge3\sqrt{x}\)
\(\Leftrightarrow\dfrac{1}{x+\sqrt{x}+1}\le\dfrac{1}{3\sqrt{x}}̸\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\le\dfrac{1}{3}\)
1)\(ĐK:x\ne0\)
\(\Rightarrow6\sqrt{x}-12-3\sqrt{x}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)
\(\Leftrightarrow\sqrt{x}>4\)
\(\Leftrightarrow x>16\)
Vậy với x>16 thì \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\)
2)Đặt =\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(\Rightarrow Ax+\left(A-1\right)\sqrt{x}+A=0\)
\(\Delta\ge0\Leftrightarrow A^2-2A+1-4A^2\ge0\)
\(\Leftrightarrow-3A^2-2A+1\ge0\)
\(\Leftrightarrow-1\le A\le\dfrac{1}{3}\)
Vậy \(A\le\dfrac{1}{3}\)
