Question

\(\dfrac{\sqrt{a+1}}{\sqrt{a^2-1}-\sqrt{a^2+a}}+\dfrac{1}{\sqrt{a-1}+\sqrt{a}}+\dfrac{\sqrt{a^3}-a}{\sqrt{a}-1}\)rut gọn giúp mik bài này nha mn,mik cảm ơn nhiều

Answer 1

\(A=\dfrac{\sqrt{a+1}}{\sqrt{a+1}.\sqrt{a-1}-\sqrt{a}.\sqrt{a+1}}+\dfrac{1}{\sqrt{a-1}+\sqrt{a}}+\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\\ =\dfrac{\sqrt{a+1}}{\sqrt{a+1}\left(\sqrt{a-1}-\sqrt{a}\right)}+\dfrac{1}{\sqrt{a-1}+\sqrt{a}}+a\\ =\dfrac{1}{\sqrt{a-1}-\sqrt{a}}+\dfrac{1}{\sqrt{a-1}+\sqrt{a}}+a\\ =\dfrac{\sqrt{a-1}+\sqrt{a}}{a-1-a}+\dfrac{\sqrt{a-1}-\sqrt{a}}{a-1-a}+a\\ =\dfrac{2\sqrt{a-1}}{-1}+a\\ =-2\sqrt{a-1}+a.\)