Question

Rút gọn biểu thức  A=\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

Answer 1

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

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

=\(\left(\frac{1-\left(\sqrt{a}\right)^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)

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

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

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

=\(1\)