Question

Se considera expresia E(x)=(2/x-1 - 1/x+1) : (x+3)(x-1), unde x este nr. Real, x≠-3, x≠-1 si x≠1. Aratati cs E(x)=1/x+1, pentru orice x numar real, x≠-3,
x≠-1 si x≠-1. VA ROG! AJUTATI-MA! /- BARA DE FRACTIE

Answer 1

E(x)= (2/x-1 - 1/x+1) : (x+3)(x-1)

(2/(x-1) - 1/(x+1))*(x-1)(x+1) 

2/x-1 - 1/x+1= 2(x+1)-(x-1)/(x+1)(x-1)


2(x+1)-(x-1)/(x+1)(x-1)*(x-1)(x+3)=
x+3/(x+1)(x-1)*(x-1)(x+3)=
1/x+1

Answer 2

Exercitiu e scris gresit, a pierdut o linie de fractie, adevarata forma este:
[tex]E(x)=(\frac{2}{x-1}- \frac{1}{x+1}): \frac{x+3}{x-1}= \frac{2x+2-x+1}{(x-1)(x+1)}* \frac{x-1}{x+3}= \frac{(x+3)(x-1)}{(x-1)(x+1)(x+3)}= \frac{1}{x+1} [/tex]