a) A=$\frac{x}{x-3}$-$\frac{x-1}{x+3}$+$\frac{5x+3}{9-x^2}$ =$\frac{x}{x-3}$-$\frac{x-1}{x+3}$-$\frac{5x+3}{(x-3)(x+3)}$ =$\frac{x^2+3x-(x^2-3x-x+3)-5x-3}{MTC}$ =$\frac{x^2+3x-x^2+3x+x-3-5x-3}{MTC}$ =$\frac{2x-6}{MTC}$ =$\frac{2(x-3)}{MTC}$ = $\frac{2}{x+3}$
b) Thay x=5 (TMĐK)
Ta có: A=$\frac{2}{x+3}$= $\frac{2}{5+3}$= $\frac{2}{8}$= $\frac{1}{4}$
Vậy x=5 thì A=$\frac{1}{4}$
c) A=$\frac{2}{7}$=>$\frac{2}{x+3}$= $\frac{2}{7}$
=>2.7=2.(x+3)
14=2x+6
14-6=2x
8=2x
=>x=4
Vậy x=4 thì A=$\frac{2}{7}$
Bạn thông cảm nha mk quên cách làm câu d r
