Question

C= \(\dfrac{32}{xy}\sqrt{\dfrac{x^4y^2}{16}}\) với y < 0

D= \(\dfrac{30}{x^2-y^2}\sqrt{\dfrac{4\left(x^2-2xy+y^2\right)}{36}}\) với x>y>0 và x ≠ y

Answer 1

\(C=\dfrac{32}{xy}\cdot\sqrt{\dfrac{x^4y^2}{16}}\)

\(=\dfrac{32}{xy}\cdot\dfrac{x^2}{4}\cdot\sqrt{y^2}\)

\(=\dfrac{8x}{y}\cdot\left|y\right|=\dfrac{8x}{y}\cdot\left(-y\right)=-8x\)

\(D=\dfrac{30}{x^2-y^2}\cdot\sqrt{\dfrac{4\left(x^2-2xy+y^2\right)}{36}}\)

\(=\dfrac{30}{\left(x-y\right)\left(x+y\right)}\cdot\sqrt{\dfrac{\left(x-y\right)^2}{9}}\)

\(=\dfrac{30}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)}{3}=\dfrac{10}{x+y}\)