Hãy so sánh \(A{\rm{ }} = {\rm{ }}2{\rm{ }}\left( {x{\rm{ }} - {\rm{ }}y} \right){\rm{ }}x{y^3}{\rm{ }}.{\rm{ }}\frac{{\sqrt {{x^2}{y^3}} {\rm{ }}}}{{\sqrt {{x^4}{y^5}{{\left( {x -...

* Hướng dẫn giải

Với y < 0 < x ⇒ x - y > 0

Ta có: |x| = x; |y| = -y; |x - y| = x - y

\(\begin{array}{l} A = 2\left( {x - y} \right)x{y^3}.\frac{{\sqrt {{x^2}{y^3}} {\rm{ }}}}{{\sqrt {{x^4}{y^5}{{\left( {x - y} \right)}^2}} }}\\ = 2\left( {x - y} \right)x{y^3}\sqrt {\frac{{{x^2}{y^3}}}{{{x^4}{y^5}{{\left( {x - y} \right)}^2}}}} \\ = 2\left( {x - y} \right)x{y^3}\sqrt {\frac{1}{{{x^2}{y^2}{{\left( {x - y} \right)}^2}}}} \\ = 2\left( {x - y} \right)x{y^3}\frac{1}{{\sqrt {{x^2}{y^2}{{\left( {x - y} \right)}^2}} }}\\ = 2\left( {x - y} \right)x{y^3}\frac{1}{{|x|.|y|.|x - y|}}\\ = \frac{{2\left( {x - y} \right)x{y^3}}}{{ - xy(x - y)}} - 2{y^2} \end{array}\)

Vậy A < 0