Question

\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

Answer 1

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

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

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

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

=\(\left(\sqrt{x}-1\right)\cdot\sqrt{x}\)

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