1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18请问这个方程式怎么

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1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18请问这个方程式怎么

1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18请问这个方程式怎么
1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18请问这个方程式怎么

1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18请问这个方程式怎么
拆项法,利用1/x(x+3)=1/3[1/x-1/(x+3)],方程化为:
1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=3/(2x+18)
1/3[1/x-1/(x+9)]=3/(2x+18)
去分母:2x+18-2x=9x
得:x=2
经检验,它是原方程的根

1/ X(x+3)+1/(x+3)(x+6)+1/(x+6)(x+9)=3/2x+18
(1/3)[1/x - 1/(x+3) +1/(x+3) -1/(x+6) +1/(x+6) -1/(x+9)]=3/(2x+18)
1/x - 1/(x+9)=9/2(x+9)
1/x=11/2(x+9)
11x=2x+18
解得x=2

3/ X(x+3)+3/(x+3)(x+6)+3/(x+6)(x+9)=9/2(x+9)
1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)=9/2(x+9)
1/x -11/2(x+9)=0
2x+18-11x=0
∴x=2

左边第一项,1/[x(x+3)]可以写成[1/3]*[1/x-1/(x+3)],同理,第二项可写成[1/3]*[1/(x+3)-1/(x+6)],第三项可写成[1/3]*[1/(x+6)-1/(x+9)]。

那么原方程左边可写成[1/3]*[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]

就是[1/3]*[1/x-1/(x+9)]

即3/[x(x+9)]=3/2x+18

这个方程会解吧

等下上图片