一道代数题,求解
来源:百度知道 编辑:UC知道 时间:2024/05/29 23:33:24
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a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
a(1/b+1/c)+1+b(1/c+1/a)+1+c(1/a+1/b)+1=0
a(1/b+1/c)+a*1/a+b(1/c+1/a)+b*1/b+c(1/a+1/b)+c*1/c=0
a(1/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)*(1/a+1/b+1/c)=0
a+b+c=0
或1/a+1/b+1/c=0
(bc+ac+ab)/(abc)=0
ab+ac+bc=0
a^2+b^2+c^2=1
a^2+b^2+c^2+2ab+2ac+2bc=1+0
(a+b+c)^2=1
a+b+c=1或-1
综上所述a+b+c=0或1或-1
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)+3=0
(b+c)/a+(a+c)/b+(a+b)/c+3=0
(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=0
(a+b+c)(1/a+1/b+1/c)=0
∴a+b+c=0或1/a+1/b+1/c=0
若1/a+1/b+1/c=0,则有:ab+ac+bc=0
∴1=a^2+b^2+c^2+2ab+2ac+2bc=(a+b+c)^2
∴a+b+c=1或a+b+c=-1
故a+b+c的值可能为0、1、-1
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3可化为
(a+b+c)(1/a+1/b+1/c)=0
(a+b+c)(ab+bc+ac)/(abc)=0
即(a+b+c)(ab+bc+ac)=0
1.当ab+bc+ac=0,则2ab+2bc+2ac=0