Question

ajutați-mă va rog...dau coronițe​

Answer 1

Răspuns:

($\frac{x-y}{x(x-y)}$ - $\frac{1}{(x-y)(x+y)}$ × $\frac{(y-x)^{2} }{x+y}$) × $\frac{(x+y)^{2} }{y^{2} }$

simplifica x-y cu x-y la prima fractie

($\frac{1}{x}$ - $\frac{1}{(x-y)(x+y)}$ × $\frac{(x-y)^{2} }{x+y}$) × $\frac{(x+y)^{2} }{y^{2} }$

simplifica x-y de la a treia fractie cu x-y de la a doua fractie

($\frac{1}{x}$ - $\frac{1}{x+y}$ × $\frac{x-y}{x+y}$) × $\frac{(x+y)^{2} }{y^{2} }$

($\frac{1}{x}$ - $\frac{x-y}{(x+y)^{2} }$) × $\frac{(x+y)^{2} }{y^{2} }$ (am inmultit fractiile)

$\frac{(x+y)^{2} - x(x-y) }{x(x+y)^{2} }$ × $\frac{(x+y)^{2} }{y^{2} }$

$\frac{(x+y)^{2} - x^{2} +xy }{x(x+y)^{2} }$ × $\frac{(x+y)^{2} }{y^{2} }$

$\frac{x^{2} +2xy+y^{2}-x^{2} +xy }{x(x+y)^{2} }$ × $\frac{(x+y)^{2} }{y^{2} }$

simplifici $(x+y)^{2}$ cu cel de la numitorul primei fractii

redu $x^{2}$ cu -$x^{2}$

$\frac{2xy+y^{2}+xy }{x}$ × $\frac{1}{y^{2} }$

$\frac{3xy+y^{2} }{x}$ × $\frac{1}{y^{2} }$

$\frac{y(3x+y)}{x}$ × $\frac{1}{y^{2} }$

simplifica $y^{2}$ cu y

$\frac{3x+y}{x}$ × $\frac{1}{y}$

$\frac{3x+y}{xy}$