Question

Answer 1

\(B=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}\)

\(=\left(\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}+3}{2\sqrt{x}-1}\)

Để B<0 thì \(\dfrac{2\sqrt{x}+3}{2\sqrt{x}-1}< 0\)

=>\(2\sqrt{x}-1< 0\)

=>\(\sqrt{x}< \dfrac{1}{2}\)

=>\(0< =x< \dfrac{1}{4}\)