Đáp án:
($\frac{x}{x+3}$ -$\frac{2}{x-3}$ +$\frac{x²-1}{9-x²}$ ):(2-$\frac{x+5}{3+x}$ )
=($\frac{x}{x+3}$ -$\frac{2}{x-3}$ +$\frac{x-1}{(3-x)(3+x)}$ ):($\frac{2(3+x)(x+5)}{3+x}$)
=($\frac{x}{x+3}$-$\frac{2}{x-3}$ +$\frac{x²-1}{-(x-3)(x+3)}$ ):$\frac{6+2x-x-5}{3+x}$
=($\frac{x}{x+3}$-$\frac{2}{x-3}$ -$\frac{x²-1}{(x-3)(x+3)}$ ): $\frac{1+x}{3+x}$
=$\frac{x(x+3)-2(x+3)-(x²-1)}{(x-3)(x+3)}$ .$\frac{3+x}{1+x}$
=$\frac{x²-3x-2x-6-x²+1}{x-3}$ .$\frac{1}{1+x}$
=$\frac{-5x-5}{x-3}$ .$\frac{1}{1+x}$
=$\frac{-5(x+1)}{x-3}$ .$\frac{1}{1+x}$
=$\frac{-5}{x-3}$
Giải thích các bước giải:
chúc bn hk tốt
