Question

Rút gọn: P= \(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1-\sqrt{a}}{1-a}\right)^2\) (với a\(\ge\) 0, a\(\ne\) 1)

Answer 1

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

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

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

\(P=1\)