∵1/X-1/(X+N)
=(X+N)/[X(X+N)]-X/[X(X+N)]
=N/[X(X+N)]

∴1/[X(X+N)]=1/N[1/X-1/(X+N)]

• 1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18)   解出來這個(gè)題，100財(cái)富給你

????????????1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=1/3[1/X-1/(X+3)+1/(X+3)-1/(X+6)+1/(X+6)-1/(X+9)]=1/3[1/X-1/(X+9)]=1/3*9/[X(X+9)]=3/(X²+9x)?????????????????????????????????????????????????????????X²+9x=2X+18 ??x²+7X-18=0???x1=2?? x2=-9(???????????)??X=2??д???????????????????????????????????1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)=3/(2X+18)1/3[1/X(X+3)+1/(X+3)(X+6)+1/(X+6)(X+9)]=3/(2X+18)[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)]=9/(2x+18)1/x-1/(x+9)=9/2(x+9)(x+9-x)/x(x+9)=9/2(x+9)9/x(x+9)=9/2(x+9)∴x(x+9)=2(x+9)x²+9x-2x-18=0x²+7x-18=0∴x+9=0 x-2=0x=-9 (不符題意，舍去)∴x=2 若還有疑問，可在HI里復(fù)制提問，免得浪費(fèi)財(cái)富。