1.已知x^2减5x减1等于0,那么x^2+1/x^2=?2.已知a=x+6,b=x+4,c=x+3.求a^2+b^2+c^2-ab-bc-ca的值 3.把（x^2-x+1)^6展开后得a12x^12+a11x^11+.+a2x^2+a1x

1 x^2+5x-1=0x+5-1/x=0x-1/x=-5两边平方得x^2+1/x^2-2=25故x^2+1/x^2=272 a-b=2 a-c=3 b-c=1a^2+b^2+c^2-ab-bc-ca=(1/2)[(a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ac)]=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]=(1/2)[2^2+1^2...