Question

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

Answer 1

ĐK: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\end{matrix}\right.\)

\(C=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\)

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

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