\frac{5}{х-у}-\frac{3}{х+y}; . Created by sabinasabina. algebra-ru

\frac{5}{х-у}-\frac{3}{х+y}; ...

sabinasabina @sabinasabina

July 2021 2 7 Report

\frac{5}{х-у}-\frac{3}{х+y};

Yena

$\frac{5}{x-y}-\frac{3}{x+y}=\frac{5(x+y)-3(x-y)}{(x+y)(x-y)}=\frac{5x+5y-3x+3y}{(x+y)(x-y)}=\frac{2x+8y}{x^2-y^2}$

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1 votes Thanks 1

ZhekaBeliy

$\frac{5x+5y-3x+5y}{(x+y)(x-y)}=\frac{2x+8y}{(x+y)(x-y)}\\ \left \{ {{2x+8y=0 | /2} \atop {(x+y)(x-y)\neq0}} \right.\\ \left \{ {{x+4y=0} \atop {(x+y)(x-y)\neq0}} \right.$

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