a² 5b² c² 4(ab-b c)-2c 5=0,2a-b c=

∵ab/(a+b)=1/3,bc/(b+c)=1/4,ca/(c+a)=1/5取倒数,得(a+b)/ab=3,(b+c)/bc=4,(c+a)/ca=5∴(ac+bc)/abc=3,(ab+ac)/a

(a-b)/ab-(a-c)/ac+(b-c)/bc　　=c(a-b)/abc-b(a-c)/abc+a(b-c)/abc　　=(ac-cb-ab+bc+ab-ac)/abc　　=0/abc　　=0

ab/(a+b)=1/3(a+b)/ab=3a/ab+b/ab=31/b+1/a=31/b+1/c=41/a+1/c=52(1/a+1/b+1/c)=121/a+1/b+1/c=6(ab+bc+ca)

ab=（a+b)/3,bc=(b+c)/4,ac=(a+c)/51/3=(a+b)/ab=1/a+1/b,1/4=1/b+1/c,1/5=1/a+1/c(1/a+1/b)+(1/b+1/c)+(1/a

ab/(a+b)=1/3(a+b)/ab=3则a/ab+b/ab=31/b+1/a=3同理1/c+1/b=41/c+1/a=5相加2(1/a+1/b+1/c)=121/a+1/b+1/c=6(ab+b

ab/(b-c)(c-a)+bc/(a-b)(c-a)+ac/(a-b)(b-c)=ab(a-b)/(a-b)(b-c)(c-a)+bc(b-c)/(a-b)(b-c)(c-a)+ac(c-a)/(a

已知的分别倒数后1/a+1/b=31/b+1/c=41/a+1/c=5三式相加除以2得:1/a+1/b+1/c=6abc/(ab+bc+ac)=1/(1/c+1/b+1/a)=1/6

ab/a+b=1/31/[1/b+1/a]=1/31/b+1/a=3bc/b+c=1/41/[1/c+1/b]=1/41/c+1/b=4ac/a+c=1/51/[1/c+1/a]=1/51/c+1/a

abc/(ab+bc+ac)=1/6.因为ab/(a+b)=1/3==>(a+b)/ab=3==>1/a+1/b=3同理:1/b+1/c=4,1/a+1/c=5.而(ab+bc+ac)/abc=1/a

ab/(a+b)=1/3取倒数(a+b)/ab=3a/ab+b/ab=31/b+1/a=3同理1/b+1/c=41/a+1/c=5相加2(1/a+1/b+1/c)=121/a+1/b+1/c=6通分(

将已知条件全部倒数,得：(a+b)/(ab)=3,(b+c)/(bc)=4,(a+c)/(ac)=5则1/a=2,1/b=1,1/c=3(ab+bc+ac)/(abc)=1/a+1/b+1/c=6所以

1、（a+b)c/abc-（b+c）/abc=（c-a)/ac2、3x/（x-3）²+x/x-3=x²/（x-3)²3、（1+x-x²）/x（1+x）4、（c&

因为AB/A+B=1/3AC/A+C=1/4BC/B+C=1/5所以A+B/AB=3A+C/AC=4B+C/BC=5A+B/AB+A+C/AC+B+C/BC=3+4+5=122(AB+AC+BC)/A

ab≠0a+b=3ab,1/a+1/b=3b+c=4bc,1/b+1/c=4c+a=5ac,1/c+1/a=51/a+1/b+1/b+1/c+1/c+1/a=3+4+5=121/a+1/b+1/c=6

ab=（a+b)/3,所以3ab=a+b,所以3=1/a+1/b（1）bc=(b+c)/4,所以4bc=b+c,所以4=1/b+1/c（2）ac=(a+c)/5,所以5ac=a+c,所以5=1/a+1

ab/(a+b)=1/3取倒数(a+b)/ab=3a/ab+b/ab=31/b+1/a=3同理1/b+1/c=41/a+1/c=5相加2(1/a+1/b+1/c)=121/a+1/b+1/c=6通分(

ab/(a+b)=1/3;bc/(b+c)=1/4;ca/(c+a)=1/51/a+1/b=31/b+1/c=41/c+1/a=5相加除以2,得1/a+1/b+1/c=6所以(a+b+c)/abc=6

3/a=4/b=5/c所以a=3c/5,b=4c/5所以(ab-bc-ac)/(a^2+b^2+c^2)=(12c^2/25-4c^2/5-3c^2/5)/(9c^2/25+16c^2/25+c^2)

c/b+c=1/4变形b+c/bc=4,ca/c+a=1/5变形为c+a/ca=5.然后可以求出1/a,1/b,1/c.最后求出1/a+1/b+1/c,1/a+1/b+1/c的倒数就是abc/ab+b