Yg bagian d mumet bantuan nya dong ka trima ksih.

---Math---

d. -x/2y × ^3Vx^3y^3 : Vx^2y^2
=> -x/2y × x^3/3 y^3/3 : x^2/2 y^2/2
=> -x/2y × xy : xy ----> kerjakan bagi dulu agar mudah (xy : xy = 1)
=> -x/2y × 1
=> -x/2y

[tex]\displaystyle (3\sqrt[3]{x^3y^6})(\sqrt{x^2y^4})=3\cdot xy^2\cdot xy^2\\\boxed{\boxed{(3\sqrt[3]{x^3y^6})(\sqrt{x^2y^4})=3x^2y^4}}\\\\\\-\frac x{2y}\cdot\sqrt[3]{x^3y^3}:\sqrt{x^2y^2}=-\frac x{2y}\cdot (xy):(xy)\\-\frac x{2y}\cdot\sqrt[3]{x^3y^3}:\sqrt{x^2y^2}=-\frac x{2y}\cdot 1\\\boxed{\boxed{-\frac x{2y}\cdot\sqrt[3]{x^3y^3}:\sqrt{x^2y^2}=-\frac x{2y}}}[/tex]