1.7.15; 1.7.16 пожалуйста. Created by stanovova00. algebra-ru

1.7.15; 1.7.16 пожалуйста...

stanovova00 @stanovova00

December 2021 1 16 Report

1.7.15; 1.7.16 пожалуйста

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NNNLLL54

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$1)\quad \frac{(x^2-y^2)^2}{x^2+2xy+y^2}:\left (\frac{1}{x^2}-\frac{1}{y^2}\right)= \frac{(x^2-y^2)^2}{(x+y)^2}:\frac{y^2-x^2}{x^2y^2}=\\\\=\frac{(x^2-y^2)^2}{(x+y)^2}\cdot \frac{x^2y^2}{-(x^2-y^2)}=-\frac{x^2y^2(x^2-y^2)}{(x+y)^2}=-\frac{x^2y^2(x-y)(x+y)}{(x+y)^2}=\\\\=-\frac{x^2y^2(x-y)}{x+y}$

$2)\quad (\frac{1}{a}+\frac{1}{b})\cdot \left ((a+b)^2-\frac{a^3-b^3}{a-b}\right )=\\\\=\frac{b+a}{ab}\cdot \left((a+b)^2-\frac{(a-b)(a^2+ab+b^2)}{a-b}\right )=\\\\=\frac{a+b}{ab}\cdot ((a+b)^2-(a^2+ab+b^2))=\\\\=\frac{a+b}{ab}\cdot (a^2+b^2+2ab-a^2-ab-b^2)=\frac{a+b}{ab}\cdot ab=a+b$

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