已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值
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![已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值](/uploads/image/z/5082628-4-8.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0a.b.c%E6%BB%A1%E8%B6%B3a%5E2%2Bb%5E2%2Bc%5E2%3D1%2C%E4%B8%94a%281%2Fb%3D1%2Fc%29%2Bb%281%2Fc%2B1%2Fa%29%2Bc%281%2Fa%2B1%2Fb%29%3D-3%2C%E6%B1%82a%2Bb%2Bc%E7%9A%84%E5%80%BC)
已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值
已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值
已知非零实数a.b.c满足a^2+b^2+c^2=1,且a(1/b=1/c)+b(1/c+1/a)+c(1/a+1/b)=-3,求a+b+c的值
∵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/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)(ab+bc+ca)/abc=0
若a+b+c=0,则问题得解.
若ab+bc+ca=0,又因为(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
故(a+b+c)^2=1+0=1
a+b+c=1或-1
解:∵a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-3
解:∵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/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)(ab+bc+ca)/abc=0
若a+b+c=0,则问题得解.
若ab+bc+ca=0...
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解:∵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/a+1/b+1/c)+b(1/a+1/b+1/c)+c(1/a+1/b+1/c)=0
(a+b+c)(ab+bc+ca)/abc=0
若a+b+c=0,则问题得解.
若ab+bc+ca=0,又因为(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)
故(a+b+c)^2=1+0=1
a+b+c=1或-1
回答者: 刘念123456
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