(x-3)/(x-2)-(x-2)/(x-1)=(x-5)/(x-4)-(x-4)/(x-3) 帮个忙呗~~~

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(x-3)/(x-2)-(x-2)/(x-1)=(x-5)/(x-4)-(x-4)/(x-3) 帮个忙呗~~~
(x-3)/(x-2)-(x-2)/(x-1)=(x-5)/(x-4)-(x-4)/(x-3) 帮个忙呗~~~

(x-3)/(x-2)-(x-2)/(x-1)=(x-5)/(x-4)-(x-4)/(x-3) 帮个忙呗~~~
1-1/(x-2)-[1-1/(x-1)]=1-1/(x-4)-[1-1/(x-3)] 1/(x-1)-1/(x-2)=1/(x-3)-1/(x-4) ((x-2)-(x-1))/((x-1)(x-2))=((x-4)-(x-3))/((x-3)(x-4)) -1/((x-1)(x-2))=1/((x-3)(x-4)) (x-3)(x-4)=(x-1)(x-2) x-7x+12=x-3x+2 10=4x x=5/2
求采纳

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原式=(x-2 1)/(X-2) - (X-3 1)/(X-3) - (X-4 1)/(X-4) (X-5 1)/(X-5) =1 1/(X-2) -1 - 1/(X-3) -1 - 1/(X-4) 1 1/(X-5) =1/(X-2) 1/(X-5) - 1/(X-3) - 1/(X-4) =(X-5 X-2)/(X-2)(X-5) - (X-4 X-3)/(X-3)(X-4) =(2X-7)/(X^2-7X 10) - (2X-7)/(X^2-7X 12) =(2X-7)[1/(X^2-7X 10) - 1/(X^2-7X 10 2) 换元；令y=X^2-7X 10,得原式=（2x-7)[ 1/y - 1/（y 2)]=(2X-7)[2/(y^2 2y)] =(4x-14)/(x-2)(x-3)(x-4)(x-5)

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