Question

Answer 1

\(Q=\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)

\(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{a\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(Q=\dfrac{1\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a\left(\sqrt{a}-1\right)\left[a-1-\left(a-2\right)\right]}=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a\left(\sqrt{a}-1\right)\left(a-1-a-2\right)}=\dfrac{\sqrt{a}-2}{-3a}\)