Question

\(\dfrac{a\sqrt{a}-2a+\sqrt{a}}{\sqrt{a}-a}\)

Answer 1

`ĐKXĐ: a >= 0`

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

Answer 2

Đk: `a >=0`.

Với `a` thỏa mãn đk thì ptr tương đương với:
`( sqrt a - 2a + sqrt a)/(sqrt a -a)`

`= (sqrt a ( a - 2 sqrt a + 1))/(sqrt a ( 1 - sqrt a))`

`= (sqrt a - 1)^2/(1-sqrt a)`

`= 1 - sqrt a.`