Question

\(\dfrac{x+y}{y}.\sqrt{\dfrac{xy^2}{x^2+2xy+y^2}}\)

\(\dfrac{1}{3-b}.\sqrt{\dfrac{9-6b+b^2}{b^2}}\left(b>3\right)\)

Answer 1

\(\dfrac{x+y}{y}.\sqrt{\dfrac{xy2}{x^2+2xy+y^2}}\) \(\Leftrightarrow\) \(\dfrac{x+y}{y}.\dfrac{\sqrt{xy^2}}{\sqrt{\left(x+y\right)^2}}\)

\(\dfrac{x+y}{y}.\dfrac{y\sqrt{x}}{x+y}\) = \(\sqrt{x}\)

đk : \(b\ne3;b\ne0\)

\(\dfrac{1}{3-b}.\sqrt{\dfrac{9-6b+b^2}{b^2}}\) \(\Leftrightarrow\) \(\dfrac{1}{3-b}.\sqrt{\dfrac{\left(3-b\right)^2}{b^2}}\)

\(\Leftrightarrow\) \(\dfrac{1}{3-b}.\dfrac{b-3}{b^2}\) (vì \(b>3\)) \(\Leftrightarrow\) \(\dfrac{b-3}{3b^2-b^3}\)