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已知abc=1,求分式v1/ab+a+1w+v1/bc+b+1w+v1/...

证明:∵abc=1∴1/ab+a+1 +1/bc+b+1 +1/ca+c+1=c/c(ab+a+1)+ ac/ac(bc+b+1) +1/(ca+c+1)=c/(abc+ac+c)+ ac/(abc+abc+ac) +1/(ca+c+1)=c/(ca+c+1)+ ac/ (ca+c+1)+ 1/(ca+c+1)=(ca+c+1)/(ca+c+1)=1

1/(ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=abc/(ab+a+abc)+1/(bc+b+1)+1/(ca+c+1)=(bc+1)/(bc+b+1)+1/(ca+c+1)=(bc+abc)/(bc+b+abc)+1/(ca+c+1)=(c+ca)/(ca+c+1)+1/(ca+c+1)=(ca+c+1)/(ca+c+1)=1

1/ab+a+1 + 1/bc+b+1 + 1/ac+c+1 =abc/(ab+a+abc)+ 1/(bc+b+1) + 1/(ac+c+1) =bc/(bc+b+1)+ 1/(bc+b+1) + 1/(ac+c+1) =(bc+1)/(bc+b+1)+ 1/(ac+c+1) =(bc+1)/(bc+b+abc)+ 1/(ac+c+1) =(bc+1)/[b(c+1+ac)]+ b/[b(ac+c+1)] =(bc+1+b)/[b(c+1+ac)] =(bc+abc+b)/(bc+b+abc) =1

因为abc=1所以1/(ab+a+1)=c/(abc+ac+c)=c/(ac+c+1)1/(bc+b+1)=ac/(abc^2+abc+ac)=ac/(ac+c+1)因此原式=c/(ac+c+1)+ac/(ac+c+1)+1/(ac+c+1)=(ac+c+1)/(ac+c+1)=1

abc=1 1/(ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=1/(ab+a+1)+a/(abc+ab+a)+ab/(abca+abc+ab)=1/(ab+a+1)+a/(1+ab+a)+ab/(a+1+ab)=(ab+a+1)/(ab+a+1)=1

(1)因为abc=1.a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)=a/(ab+a+1)+ab/(abc+ab+a)+abc/(abac+abc+ab)=a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab)=(ab+a+1)/(ab+a+1)=1(2)a1/b+1/c+b1/a+1/c+c1/a+1/b =ab+c/bc+ba+c/ac+ca+b/

解,用整体法:a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1)等价于1/(b+1+bc)+1/(c+1+ac)+1/(a+1+ab)(分子,分母同除分子) 等价于ac/(1+ac+c)+ab/(1+ab+a)+bc/(1+cb+b)(分子,分母同乘分母能构成abc的项的差那个数) 3*[a/(ab+a+1)+b/(bc+b+1)+c/(

[a/(ab+a+1)]+[b/(bc+b+1)]+[c/(ac+c+1)] =abc/(ab*bc+abc+bc) + abc/(bc*ac+abc+ac) + abc/(ac*ab+abc+ab)=1/(b+1+bc) + 1/(c+1+ac) + 1/(a+1+ab)=1/(ab+a+1) + 1/(bc+b+1) + 1/(ac+c+1) 与原式比较得a=b=c=1所以[a/(ab+a+1)]+[b/(bc+b+1)]+[c/(ac+c+1)]=1

分子是1 你abc=1 你带入就可以了 只能化简求不出答案 真的 不信你可以带入数值 假设A,B. C的值 A=1B=1C=1时是9 A=0.5B=1C=2是10

a^2+2ab+b^2=(a+b)^2 ab^2+a^2*b=ab(a+b) 所以所求分式=(a+b)^2/ab(a+b)=(a+b)/ab 因为1/a+1/b=(a+b)/ab=所要求的分式 所以该分式=5

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