（x2+ax+b)(x2-3x+4) =x^4-3x^3+4x^2+ax^3-3ax^2+4ax+bx^2-3bx+4b =x^4-(3-a)x^3+(4-3a+b)x^2+(4a-3b)+4b 3-a=0 => a=3 4-3a+b=0 4-3*3+b=0 b=5 设任取0-9中3个数X,Y,Z. 6个两位数分别是：10X+Y,10Y+X,10X+Z,10Z+X,1...