原式=1/(x+1)*[(x-3)/(x+3)²+2(x+3)/(x+3)²]-x/(x+3)(x-3)=1/(x+1)*[(x-3+2x+6)/(x+3)²]-x/(x+3)(x-3)=1/(x+1)*[3(x+1)/(x+3)²]-x/(x+3)(x-3)=3/(x+3)²-x/(x+3)(x-3)=(3x-9-x²-3x)/(x+3)²(x-3)=-(x²+9))/(x+3)²(x-3)
