Question

解方程：1/（X的平方-5X+6）-1/（x的平方-4X+3）+1/（x的平方-3x+2）=1/x-1

已知：n（n+1）分之一=n分之一 -（n+1）分之一

Answer 1

1/（X的平方-5X+6）-1/（x的平方-4X+3）+1/（x的平方-3x+2）=1/x-1
1/(x-3)（x-2）-1/(x-3)(x-1）+1/（x-2）(x-1)=1/x-1
x-1-(x-2)+x-3=X的平方-5X+6
x-2=X的平方-5X+6
所以
x²-6x+8=0
(x-2)(x-4)=0
x=2或x=4
经检验，得x=2是增根，所以
方程的解为x=4

Answer 2

1/(x^2-5x+6) -1/(x^2-4x+3) +1/(x^2-3x+2) = 1/(x-1)
1/[(x-2)(x-3)] - 1/[(x-1)(x-3)]+1/[(x-1)(x-2)] = 1/(x-1)
[-(x-1)+(x-2)-(x-3)+(x-2)(x-3)]/[(x-1)(x-2)(x-3)]=0
(x^2-6x+8)/[(x-1)(x-2)(x-3)]=0
(x-2)(x-4)/[(x-1)(x-2)(x-3)] =0
(x-4)/[(x-1)(x-3)]=0
x=4

Answer 3

1/（X的平方-5X+6）-1/（x的平方-4X+3）+1/（x的平方-3x+2）=1/x-1
1/[（x-2)(x-3)]-1[(x-1)(x-3)]+1[(x-1)(x-2)]=1/(x-1) 分母分解因式
[1/(x-2)-1/(x-3)]-1/2[1/(x-1)-1/(x-3)]+[1/(x-1)-1/(x-2)]=1/(x-1) 拆项
1/2[1/(x-1)-1/(x-3)]=1/(x-1) 去括号，合并同类项
1[(x-1)(x-3)]=1/(x-1) 差化积
1/(x-3)=1 等式两边同除1/（x-1）
x-3=1
x=4

Answer 4

1/(x-2)*(x-3)-1/(x-3)*(x-1)+1/(x-1)*(x-2)
=1/(x-3) -1/(x-2) -1/(x-3)*(x-1)+1/(x-2) -1/(x-1)
=1/(x-3)-1/(x-1)-1/(x-3)*(x-1)
=1/(x-1)
可得1/(x-3)-1/(x-3)*(x-1)=0
[(x-1)-(x-3)*(x-1)]/(x-3)*(x-1)=0
即x-1-x*x+4x-3=0
x*x-5x+4=0(x!=1或3)
所以x=4