分式通分(初二)1.a+b/(a-b)(b-c),b+c/(b-c)(b-a);2.1/(a-b)(a-c),1/(b-c)(b-a),1/(c-a)(c-b);2 2 23.已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bx+b/ca+c/ab-1/a-1/b-1/c;
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![分式通分(初二)1.a+b/(a-b)(b-c),b+c/(b-c)(b-a);2.1/(a-b)(a-c),1/(b-c)(b-a),1/(c-a)(c-b);2 2 23.已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bx+b/ca+c/ab-1/a-1/b-1/c;](/uploads/image/z/2694254-14-4.jpg?t=%E5%88%86%E5%BC%8F%E9%80%9A%E5%88%86%EF%BC%88%E5%88%9D%E4%BA%8C%EF%BC%891.a%2Bb%2F%28a-b%29%28b-c%29%2Cb%2Bc%2F%28b-c%29%28b-a%29%3B2.1%2F%28a-b%29%28a-c%29%2C1%2F%28b-c%29%28b-a%29%2C1%2F%28c-a%29%28c-b%29%3B2+2+23.%E5%B7%B2%E7%9F%A5a%2Bx%3D2003%2Cb%2Bx%3D2004%2Cc%2Bx%3D2005%2C%E4%B8%94abc%3D6012%2C%E6%B1%82a%2Fbx%2Bb%2Fca%2Bc%2Fab-1%2Fa-1%2Fb-1%2Fc%3B)
分式通分(初二)1.a+b/(a-b)(b-c),b+c/(b-c)(b-a);2.1/(a-b)(a-c),1/(b-c)(b-a),1/(c-a)(c-b);2 2 23.已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bx+b/ca+c/ab-1/a-1/b-1/c;
分式通分(初二)
1.a+b/(a-b)(b-c),b+c/(b-c)(b-a);
2.1/(a-b)(a-c),1/(b-c)(b-a),1/(c-a)(c-b);
2 2 2
3.已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bx+b/ca+c/ab-1/a-1/b-1/c;
分式通分(初二)1.a+b/(a-b)(b-c),b+c/(b-c)(b-a);2.1/(a-b)(a-c),1/(b-c)(b-a),1/(c-a)(c-b);2 2 23.已知a+x=2003,b+x=2004,c+x=2005,且abc=6012,求a/bx+b/ca+c/ab-1/a-1/b-1/c;
,是什么呀?要是+的话:(1):(a+b)/(a-b)(b-c)+(b+c)/(b-c)(b-a)= [(a+b)-(b+c)]/(a-b)(b-c)=(a-c)/(a-b-)(b-c) (2):1/(a-b)(a-c)+1/(b-c)(b-a)+1/(c-a)(c-b)=[(b-c)-(a-c)+(a-b)]/(a-b)(b-c)(a-c)=0 (3)a/bx应该是a/bc吧?由题意知c=b+1=a+2;a/bc+b/ac+c/ab-1/a-1/b-1/c=[(a^2+b^2+c^2)-(ab+bc+ac)]/abc=[(a-b)^2+(b-c)^2+(a-c)^2]/2abc=(1^2+1^2+2^2)/(2*6012)=1/2004
那符号写清楚点,人家连问题都看不懂,怎么答你的题啊