1.已知x^2-4x+1=0,则x^4+1/x^4=2.如果方程a/(x-2)+3=(1-x)/(1+x)有增根,那么a=3.分解因式(1)[（a^2-a-6)/(a-3)-5/(a-2)]/[(3-a)/(2a-4)(2)[1/x-(x-3)/(1-x)+2/(x^2-x)]/(1+3/x-4/x^2)(3)1/x-1/(x-1)[(x^2-1)/x-x+1](4)a^2+a+1-a^3/(a-1)(

Question

1.已知x^2-4x+1=0,则x^4+1/x^4=_______
2.如果方程a/(x-2)+3=(1-x)/(1+x)有增根,那么a=______
3.分解因式(1)[（a^2-a-6)/(a-3)-5/(a-2)]/[(3-a)/(2a-4)
(2)[1/x-(x-3)/(1-x)+2/(x^2-x)]/(1+3/x-4/x^2)
(3)1/x-1/(x-1)[(x^2-1)/x-x+1]
(4)a^2+a+1-a^3/(a-1)
(5)a^2-2a+4-(a^3+a+6)/(a+2)
(6)3x/(x^2-y^2)+2x/(y^2-x^2)+y/(x+y)(x-y)
(7)(2+3x)/(4-9x)-(1+3x)/(2-3x)-2/(3x-2)

Answer 1

1.194 2.第二题怎么算都不合理
3.(1)[（a^2-a-6)/(a-3)-5/(a-2)]/[(3-a)/(2a-4)=-2a-6
(2)[1/x-(x-3)/(1-x)+2/(x^2-x)]/(1+3/x-4/x^2)=x/(x+4)
(3)1/x-1/(x-1)[(x^2-1)/x-x+1]=(x-2)/(x-1)
(4)a^2+a+1-a^3/(a-1)=1/(1-a)
(5)a^2-2a+4-(a^3+a+6)/(a+2)=(2-a)/(2+a)
(6)3x/(x^2-y^2)+2x/(y^2-x^2)+y/(x+y)(x-y)=1/(x+y)
(7)(2+3x)/(4-9x)-(1+3x)/(2-3x)-2/(3x-2) 少抄平方了吧