Question

\(\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\cdot\frac{1+2\sqrt{a}+a}{\left(1-a\right)^2}+\sqrt{a}\) (a≥0; a≠ -1)

a ) rút gọn

Answer 1

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

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

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

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

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

=\(\sqrt{a}\)