Question

với a > 0, b > 0 thì \(\sqrt{\dfrac{a}{b}}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}\)bằng:

a) 2

b) \(\dfrac{2\sqrt{ab}}{b}\)

c) \(\sqrt{\dfrac{a}{b}}\)

d) \(\sqrt{\dfrac{2a}{b}}\)

Answer 1

\(=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a^2b}{b^2a}}=\dfrac{\sqrt{ab}}{b}+\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{ab}}{b}+\dfrac{\sqrt{ab}}{b}=\dfrac{2\sqrt{ab}}{b}\left(B\right)\)

Answer 2

B

Answer 3

B