Question

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

rút gọn biểu thức 

Answer 1

ĐK: \(a\ge0;a\ne1\)

Biểu thức trở thành:

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