已知X (A-B)=Y (B-C)=Z (C-A),求X Y Z

直接代入可得（2a-b-c）(b-c)+（2b-c-a）(c-a)+（2c-a-b）(a-b)=2a(b-c)-(b+c)(b-c)+2b(c-a)-(c+a)(c-a)+2c(a-b)-(a+b)(

不对吧gaojing198201做的对有点特殊了任意的x,y不能满足(x-c)(c-y)(y-x)=1(x-a)(a-y)(y-x)=1(x-b)(b-y)(y-x)=1这些式子的.当x=a,y=b或

令a/(x-y)=b/(y-z)=c/(z-x)=ma=m(x-y)b=m(y-z)c=m(z-x)a+b+c=m(x-y)+m(y-z)+m(z-x)=m(x-y+y-z+z-x)=m*0=0

x+y+z=2a-b-c+2b-a-c+2c-a-b=0(b-c)x+(c-a)y+(a-b)z①=a(z-y)+b(x-z)+c(y-z)=a(x+2z)+b(y+2x)+c(z+2y)=(a+2b

因为x/x=1,所以x/x+y=a,1+y=ax/x+y=c,1+y=c到底1+y是等于a还是等于c?

设：x/b+c-a=y/c+a-b=z/a+b-c＝kx=k(b+c-a)y=k(c+a-b)z=k(a+b-c)三式相加得：x+y+z=k(a+b+c)(b-c)x+(c-a)y+(a-b)z=k[

令x/(a+2b+c)=y/(a-c)=z/(a-2b+c)=k所以x=ak+2bk+ck.1y=ak-ck.2z=ak-2bk+ck.3先把a作为等式的左边独立出来通过1式得到a=(2bk+ck)/

设x+y=zkx+z=yky+z=xk将上式相加得2（x+y+z）=（x+y+z）k（2-k）（x+y+z）=0因为x+y+z不等于0所以2-k=0k=2k带回去则2（x+y-z）=y+x+2z-x-

因为x/(a-b)=y/(b-c)=z/(c-a)可得x=y(a-b)/(b-c)z=y(c-a)/(b-c)x+y+z=y(a-b)/(b-c)+y+y(c-a)/(b-c)=y(a-b+c-a)/

设X/(A-B)=Y/(B-C)=Z/(C-A)=k所以X=(A-B)k,Y=(B-C)k,Z=(C-A)k所以X+Y+Z=(A-B)k+(B-C)k+(C-A)k=(A-B+B-C+C-A)k=0×

a,b互为相反数=>a+b=0=>那么b=-a,x=4(a-2)-[a-3*(-a)]=-8.因为c、d互为倒数=>cd=1=>y=c（cd)-cd-c+1=c*1-1-c+1=0x+y=-8+0=-

设x/(a-b)=y/(b-c)=z/(c-a)=t则x+y+z=(a-b)t+(b-c)t+(c-a)t=(a-b+b-c+c-a)t=0

令a-b/x=b-c/y=c-a/z=K所以x=a-b/k,y=b-c/k,z=c-a/kx+y+z=(a-b+b-c+c-a)/k=0

(b－c)x＋(c－a)y＋(a－b)z＝bx－cx＋cy－ay＋az－bz＝az－ay＋bx－bz＋cy－cx＝a(z－y)＋b(x－z)＋c(y－x)z－y＝2c－a－b－2b＋c＋a＝3c－3b

[x,y]=a*b*c(x,y)=0

设a/2=b/7=c/5=k,则a=2k,b=7k,c=5k,x=(a+b+c)分之b=1/2y=b分之a+b=9/7z=a分之(a+b-c)=2,所以z>y>x

设a/2=b/7=c/5=k(k≠0)a=2k,b=7k,c=5kx=b/(a+b+c)=7k/(2k+7k+5k)=1/2y=(a+b)/c=(2k+7k)/5k=9/5z=(a+b+c)/a=(2

1.设x/a-b=y/b-c=z/c-a=k,x=ka-kb,y=kb-kc,z=kc-ka,三式相加,x+y+z=ka-kb+kc-ka+kc-ka=0,x+y+z=02,a（1/b+1/c）+b（

设k=(a-b)/x=(b-c)/y=(c-a)/z,则x=(a-b)/k,y=(b-c)/k,z=(c-a)/kx+y+z=(a-b+b-c+c-a)/k=0

a=X/X+Y?应为a=X/Y+Z吧!a/a+1=(X/(Y+Z))/((X/(Y+Z))+1)=X/X+Y+Zb/b+1=Y/X+Y+Zc/c+1=Z/X+Y+Z