Question

Am nevoie de ajutor va rog​

Answer 1

Răspuns:

$(\frac{x+1}{2x+1})^(^-^1^)+(\frac{2}{x+1}-\frac{1}{x+2}):\frac{x^2+x-6}{x^2-4}$

$x^2+x-6=x^2-2x+3x-6 = x(x-2)+3(x-2) <=> (x+3)(x-2)$

$x^2-4 = (x-2)(x+2)$

Inlocuim in ecuatie:

$(\frac{2x+1}{x+1})+(\frac{2}{x+1}*\frac{x+2}{x+2}-\frac{1}{x+2}*\frac{x+1}{x+1})*\frac{(x-2)(x+2)}{(x+3)(x-2)}$

$\frac{2x+1}{x+1}+\frac{2x+4-x-1}{(x+1)(x+2)} *\frac{(x+2)(x-2)}{(x+3)(x-2)}$

$\frac{2x+1}{x+1}+\frac{x+3}{(x+1)(x+2)}+\frac{(x+2)(x-2)}{(x+3)(x-2)}$ Observatie! Vom simplifica (x+2), (x-2) si (x+3)

$\frac{2x+1}{x+1}+\frac{1}{x+1}=\frac{2x+1+1}{x+1} <=> \frac{2x+2}{x+1} <=> \frac{2(x+1)}{x+1} <=> 2$

$E(x) = 2$