因?yàn)閒(x)=a*x^2+b*x
所以f(-x+5)=a*(-x+5)^2+b*(-x+5)=a*(x^2-10x+25)-bx+5b=ax^2-10ax+25a-bx+5b
f(x-3)=a*(x-3)^2+b*(x-3)=ax^2-6ax+9a+bx-3b
因?yàn)閒(-x+5)=f(x-3)
即ax^2-10ax+25a-bx+5b=ax^2-6ax+9a+bx-3b
化簡(jiǎn)得ax^2-(10a+b)x+(25a+5b)=ax^2-(6a-b)x+(9a-3b)
由上式根據(jù)待定系數(shù)法可得-(10a+b)=-(6a-b)和+(25a+5b)=(9a-3b)
即4a=-2b和16a=-8b,都推出a=-0.5b……①
又因?yàn)閒(x)=x,即a*x^2+b*x=x,得a*x^2+(b-1)x=0
方程f(x)=x有等根,得出△=(b-1)^2-4a*0=0,即△=(b-1)^2=0
所以b=1,由①式a=-0.5b可得,a=-0.5
所以f(x)=-0.5x^2+x
(2)假設(shè)存在實(shí)數(shù)m,n(m