Question

Rút gọn A=\(\dfrac{6xy^2}{x^2-y^2}\sqrt{\dfrac{x^2-2xy+y^2}{9x^2y^4}}\)với x<y<0

Answer 1

\(A=\dfrac{6xy^2}{x^2-y^2}\sqrt{\dfrac{x^2-2xy+y^2}{9x^2y^4}}\left(x< y< 0\right)\)

\(A=\dfrac{6xy^2}{\left(x-y\right)\left(x+y\right)}\sqrt{\dfrac{\left(x-y\right)^2}{\left(3xy^2\right)^2}}\)

\(A=\dfrac{6xy^2}{\left(x-y\right)\left(x+y\right)}.\left|\dfrac{x-y}{3xy^2}\right|\)

\(A=\dfrac{6xy^2}{\left(x-y\right)\left(x+y\right)}.\dfrac{x-y}{3xy^2}\)

\(=\dfrac{6xy^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).3xy^2}=\dfrac{2}{x+y}\)