Question

Answer 1

a)=(\(\dfrac{1.\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)}\)-\(\dfrac{2}{\sqrt{x}.\left(\sqrt{x}-1\right)}\)) : \(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2}\)

=\(\dfrac{\sqrt{x}-2}{\sqrt{x}.\left(\sqrt{x}-1\right)}\). \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}\)

=\(\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

Answer 2

b) Vì \(\sqrt{x}\) > 0

\(\Rightarrow\) Để P < 0 thì \(\sqrt{x}-1\) < 0

\(\Leftrightarrow\) \(\sqrt{x}\) < 1

\(\Rightarrow\)x < 1

Vậy để P < 0 thì x < 1