Question

\(\sqrt{1-a}+\sqrt{1-a^2}\)

Answer 1

đk \(a\le1\)

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

Answer 2

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