Question

cine ma ajuta și pe mine , va rog !! Exercițiul 5 ...Urgent!

Answer 1

$\displaystyle \it E(x)=\left(1+ \frac{2-x}{x+1} \right): \frac{x-1}{(2x+1)^2-(x+2)^2} \\ \\ E(x)= \frac{x+1+2-x}{x+1}: \frac{x-1}{(2x)^2+2 \cdot 2x \cdot 1+1^2-(x^2+2 \cdot x \cdot 2+2^2)} \\ \\ E(x)= \frac{3}{x+1} : \frac{x-1}{4x^2+4x+1-(x^2+4x+4)} \\ \\ E(x)= \frac{3}{x+1} : \frac{x-1}{4x^2+4x+1-x^2-4x-4} \\ \\ E(x)= \frac{3}{x+1} : \frac{x-1}{3x^2-3}\\ \\E(x)=\frac{3}{x+1} : \frac{x-1}{3(x^2-1)} \\ \\ E(x)= \frac{3}{x+1} \cdot \frac{3(x^2-1)}{x-1} \\ \\ E(x)= \frac{9(x^2-1)}{(x^2-1)} \\ \\ E(x)=9$