分解因式计算题

1.[a^2-b^2}{x^2-1}=4abx{a的绝对值不等于b的绝对值}

(a+b)(a-b)x^2-4abx-(a+b)(a-b)=0

[(a+b)x+(a-b)][(a-b)x-(a+b)]=0

x1=-(a-b)/(a+b)

x2=(a+b)/(a-b)

2.x^2-x+1=6/x^2-x

(x^2-x+1)(x^2-x)=6

(x^2-x)^2+(x^2-x)-6=0

(x^2-x+3)(x^2-x-2)=0

(x^2-x+3)(x-2)(x+1)=0

∵x^2-x+3>0

所以：x1=2,x2=-1

问题补充：2x/[x-1]+[x-1]/x=4

2x^2+[x-1]^2=4x[x-1]

2x^2+x^2-2x+1=4x^2-4x

x^2-2x-1=0

[x-1]^2=2

[(x-1)+√2][(x-1)-√2]=0

x1=1-√2

X2=1+√2

1.展开移项得(ax)^2-(bx)^2-a^2+b^2-4abx=0
[(ax)^2-2abx+b^2]-[(bx)^2+2abx+a^2]=0
(ax-b)^2-(bx+a)^2=0
(ax-b+bx+a)(ax-b-bx-a)=0
[(a+b)x+(a-b)][(a-b)x-(a+b)]=0
x=-(a-b)/(a+b)或x=(a+b)/(a-b)

2.令x^2-x=t,
t+1=6/t
t^2+t-6=0
(t+3)(t-2)=0
t=-3或t=2
则x^2-x=-3或x^2-x＝2
第一个方程Δ<0,无解
第二个方程x^2-x-2=02
(x-2)(x+1)=0
得x=2或x=-1

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