Question

Rút gọn

a)\(\dfrac{4}{a-3}.\sqrt{\dfrac{27.\left(a-3\right)^2}{48}}\)với a<3

b)\(\dfrac{x-y}{b}.\sqrt{\dfrac{ab^4}{x^2-2xy+y^2}}\)với b≠0,x<y<0

Answer 1

a) \(\dfrac{4}{a-3}.\sqrt{\dfrac{27.\left(a-3\right)^2}{48}}\)

\(=\dfrac{4}{a-3}.\dfrac{3\left|a-3\right|}{4}\)

\(=\dfrac{-3\left(a-3\right)}{a-3}\) (do a < 3)

\(=-3\)

b) \(\dfrac{x-y}{b}.\sqrt{\dfrac{ab^4}{x^2-2xy+y^2}}\)

\(=\dfrac{x-y}{b}.b^2.\sqrt{\dfrac{a}{\left(x-y\right)^2}}\)

\(=\dfrac{b\left(x-y\right).\sqrt{a}}{\left|x-y\right|}\)

\(=\dfrac{b\sqrt{a}\left(x-y\right)}{-\left(x-y\right)}\) (do \(x< y\))

\(=-b\sqrt{a}\)