x^2/(x+1)-ax-b=(x^2-ax^2-ax-bx-b)/(x+1)=[(1-a)x^2-(a+b)x-b]/(x+1)={[(1-a)x-(b+1)](x+1)+1}/(x+1)=)=(1-a)x-(b+1)+1/(x+1)
当x趋向无穷,原式=(1-a)x-(b+1)=0
所以a=1,b=-1