题已知函数f(x)=ax x+b(a≠0)满足f(2)=1,且方程f(x)=x仅有一解,求其解析式.原解:由f(2)=1,得2a+b=2,由方程f(x)=ax x+b=x,得ax2+(b-1)x=0,由已知方程仅一解知△-(b-1)2=0,联立两条件解得a=12,b=1,故f(x)=x2+x2为所求.解答错了!错在哪里?错因:上述解答过程中方程①:f(x)=axx+b= The known function f(x)=ax x+b(a ≠ 0) satisfies f(2)=1, and the equation f(x)=x has only one solution, find its analytical solution. The original solution: by f(2) ) = 1, get 2a + b = 2, from the equation f (x) = ax x + b = x, get ax2 + (b-1) x = 0, known by the equation only a solution △-(b-1 ) 2 = 0, the two conditions of the simultaneous solution a = 12, b = 1, so f (x) = x2 + x2 for the required. The solution is wrong! Where is the wrong? Cause: In the above solution process equation 1: f(x)=axx+b=