设（2-x）的5次方=ax的五次方+bx四次方+cx三次方+dx二次方+ex+f，求

（2-x）^5=(x^2-4x+4)^2(2-x) =(x^4+16x^2+16-8x^3+8x^2-32x)(2-x) =2x^4+32x^2+32-16x^3+16x^2-64x-x^5-16x^3-16x+8x^4-8x^3+32x^2 =-x^5+10x^4-40x^3+80x^2-80x+32 f=32 a+b+c+d+e+f=-1+10-40+80-80+32 =1