解：[(x-2)/(x+2)]+[4/(x²-4)]=1[(x-2)²/(x+2)(x-2)]+[4/(x²-4)]=1[(x²-4x+4)/(x²-4)]+[4/(x²-4)]=1(x²-4x+4+4)/(x²-4)=1(x²-4x+8)/(x²-4)=1x²-4x+8=x²-4-4x=-4-8-4x=-12x=3检验：把x=3代入x²-4=3²-4=5≠0所以原方程的根是x=3
