Question

Se considera expresia E(x)=$\frac{x}{x^2+3x}$-($\frac{1}{x-3}$-$\frac{1}{x+3}$) : $\frac{6}{x-3}$ , unde x este numar real, x$\neq {3}$

Answer 1

$E(x)= \frac{x}{x^2+3x} - ( \frac{x+3-x+3}{(x-3)(x+3)} )*\frac{x-3}{6} \\E(x)=\frac{x}{x^2+3x}-\frac{6}{(x-3)(x+3)} *\frac{x-3}{6} \\E(x)=\frac{x}{x(x+3)}-\frac{1}{x+3} \\E(x)=\frac{x}{x(x+3)}-\frac{x}{x(x+3)} \\E(x)=0$