B=($\frac{3}{x-3}$-$\frac{6x}{9-x²}$+ $\frac{x}{x+3}$)($1$- $\frac{2}{x+3}$)
=[$\frac{3(x-3)}{(x-3)(x+3)}$+ $\frac{6x}{(x-3)(x+3)}$+ $\frac{x(x-3)}{(x-3)(x+3)}$]( $\frac{x+3}{x+3}$-$\frac{2}{x+3}$)
=[$\frac{3x+9+6x+x²-3x}{(x-3)(x+3)}$]($\frac{x+3-2}{x+3}$)
=[$\frac{x²+6x+9}{(x-3)(x+3)}$] $\frac{x+1}{x+3}$
=$\frac{(x+3)²}{(x-3)(x+3)}$ $\frac{x+1}{x+3}$
=$\frac{x+1}{x-3}$
Chúc bạn học tốt nhé! ^^
