f（x）=ax+b/(x^+1)=yyx^2-ax+y-b=0判别：a^2-4y(y-b)>=0y^2-yb-a^2/4<=04,-1为方程y^2-yb-a^2/4=0两根b=4-1,-a^2/4=4*(-1)b=3,a=4或a=-4a+b=7或a+b=-1
