4a²+3c-3ac-4a

∵3b＝a＋2c∴（3b）²＝（a＋2c）²9b²＝a²＋4ac＋4c²∴a²－9b²＋4c²＝a²－（a&s

应是已知a、b、c为实数,且ab/a+b=1/3,bc/a+c=1/4,ac/a+c=1/5,那么abc/ab+bc+ac的值是多少?因为ab/(a+b)=1/3,bc/(b+c)=1/4,ca/(c

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)

((a^2+ab)/(-c))^3/((a^2-b^2)/(-ac))^4=-a(a+b)/c^3÷[(a+b)(a-b)/a^4c^4]=-a÷[(a+b)/a^4c]=-a^5c/(a+b)如果本

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

1/2a^2-5ac-(3a^2c-a^2b)+(3ac-4a^2c)=1/2a^2-5ac-3a^2c+a^2b+3ac-4a^2c=1/2a^2-2ac-7a^2c+a^2

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

由(a+b-c)/ac+(a+c-b/ac)+(b+c-a/bc)=1/4两边都乘以abc,得2(ab+bc+ac)-（a²+b²+c²）=abc/4-(a+b+c)&#

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)=1/3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(

因为a=-1,c=2a,b=|c|所以a=-1,c=-2,b=2原式=1/2a²-5ac-2(3a²c-a²b)+3(3ac+4a²c)=1/2a²-

5a/3(b^2)c-7b/6ac^2+11c/4(a^2)b=(5a*4a^2c-7b*2ab^2+11c*3bc^2)/(12a^2b^2c^2)=(20a^3c-14ab^3+33bc^3)/(

倒过来,就有(a+b)/ab=3,1/a+1/b=3----(1)(b+c)/bc=4,1/b+1/c=4----(2)(a+c)/ac=5,1/a+1/c=5----(3)(1)+(2)+(3),2

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

a^2-9b^2+4c^2+4ac=a^2+4c^2+4ac-9b^2=(a+2c)^2-9b^2=0

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(

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)

2

a-b+c=o-------1式,4a+2b+c=-3----2式,4ac-b平方/4a=-4-----3式2式-1式得：a+b=-1;2式-1式*4,并化简得：2b-c=-1;1式*2+2式,并化简