Question

alguem me ajuda nessas contas 12y⁵ : 4y³ =

(-2a¹xy) : (+3a²x) =

(20a⁴b²) : (-5ab) =

Answer 1

$\dfrac{12y^5}{4y^3} = \dfrac{-2a^1xy}{3a^2x} \\ \\ \\ \dfrac{12y^5}{4y^3} = \dfrac{-2axy}{3a^2x} \\ \\ \\ 3y^2 = -\dfrac{2y}{3a} \\ \\ \\ 3y^2 + \dfrac{2y}{3a} = 0 \\ \\ \\ \dfrac{9ay^2 + 2y}{3a} \\ \\ \\ => 9ay^2 + 2y$

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$\dfrac{(20a^4b^2)}{(-5ab)} \\ \\ \\=> - 4a^3b$