设关于x的多项式为f(x):f(x)-1 = p(x)*2(x+1)f(x)+2 = q(x)*3(x-2)因为x^2-x-2=（x+1）（x-2）所以(f(x)-1)*(x-2) = 2p(x)*(x+1)*(x-2)(f(x)+2)*(x+1) = 3q(x)*(x-2)*(x+1)两式相减得-3f(x)-3x = [2p(x)-3q(x)]*(x-2...