y=(2x+1)/(x-2)则，y'=[(2x+1)'·(x-2)-(2x+1)·(x-2)']/(x-2)²=[2·(x-2)-(2x+1)·1]/(x-2)²=(2x-4-2x-1)/(x-2)²=-5/(x-2)²所以，y'(1)=-5则在点(1,-3)处切线方程为：y-(-3)=(-5)·(x-1)即：5x+y-2=0
