(x+3y/x^2-3xy^2 + 3/x^2+3xy). Created by dmitrievadasha1. matematika-ru

x^2+3xy)...

July 2022 1 18 Report

(x+3y/x^2-3xy^2 + 3/x^2+3xy)

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skvrttt $\frac{x+3y}{x^2y-3xy^2}+\frac{3}{x^2+3xy}=\frac{(x+3y)^2}{xy(x-3y)(x+3y)}+\frac{3y(x-3y)}{xy(x+3y)(x-3y)}=\frac{x^2+9xy}{xy(x-3y)(x+3y)}=\\\frac{x+9y}{y(x-3y)(x+3y)}$

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dmitrievadasha1 Я пропустила у.... Извиняюсь. В знаменателе первой дроби должно быть x^2y- 3xy^2

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