Question

Answer 1

3: \(\frac{2\sqrt{x}+2}{\sqrt{x}-1}\ge-1\)

=>\(\frac{2\sqrt{x}+2}{\sqrt{x}-1}+1\ge0\)

=>\(\frac{2\sqrt{x}+2+\sqrt{x}-1}{\sqrt{x}-1}\ge0\)

=>\(\frac{3\sqrt{x}+1}{\sqrt{x}-1}\ge0\)

=>\(\sqrt{x}-1>0\)

=>\(\sqrt{x}>1\)

=>x>1

4: \(\frac{x-\sqrt{x}+4}{3\sqrt{x}-2}<1\)

=>\(\frac{x-\sqrt{x}+4-3\sqrt{x}+2}{3\sqrt{x}-2}<0\)

=>\(\frac{x-4\sqrt{x}+6}{3\sqrt{x}-2}<0\)

mà \(x-4\sqrt{x}+6=\left(\sqrt{x}-2\right)^2+2>0\forall x\) thỏa mãn ĐKXĐ
nên \(3\sqrt{x}-2<0\)

=>\(3\sqrt{x}<2\)

=>\(\sqrt{x}<\frac23\)

=>\(0\le x<\frac49\)