Question

Rút gọn biểu thức :

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

( Với a lớn hơn hoặc 0 ; a khác 1 )

Answer 1

\(A=\left(\dfrac{1-a\sqrt{a}}{1-a\sqrt{a}}+\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\left(dkxd:a\ge0,a\ne1\right)\)

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

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

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

\(=\dfrac{1-\sqrt{a}}{1-a}\) 

Vậy \(A=\dfrac{1-\sqrt{a}}{1-a}\) với \(a\ge0,a\ne1\)