Question

Giúp mình bài 1 với

Answer 1

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