已知a、b、c、是三个不等于零的有理数.1、若a+b+c=1,a^2+b^2+c^2=1,求1/a+1/b+1/c2、若a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3,求a+b+c的值
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![已知a、b、c、是三个不等于零的有理数.1、若a+b+c=1,a^2+b^2+c^2=1,求1/a+1/b+1/c2、若a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3,求a+b+c的值](/uploads/image/z/2535540-60-0.jpg?t=%E5%B7%B2%E7%9F%A5a%E3%80%81b%E3%80%81c%E3%80%81%E6%98%AF%E4%B8%89%E4%B8%AA%E4%B8%8D%E7%AD%89%E4%BA%8E%E9%9B%B6%E7%9A%84%E6%9C%89%E7%90%86%E6%95%B0.1%E3%80%81%E8%8B%A5a%2Bb%2Bc%3D1%2Ca%5E2%2Bb%5E2%2Bc%5E2%3D1%2C%E6%B1%821%2Fa%2B1%2Fb%2B1%2Fc2%E3%80%81%E8%8B%A5a%5E2%2Bb%5E2%2Bc%5E2%3D1%2Ca%281%2Fb%2B1%2Fc%29%2Bb%281%2Fa%2B1%2Fc%29%2B%281%2Fa%2B1%2Fb%29%3D+-3%2C%E6%B1%82a%2Bb%2Bc%E7%9A%84%E5%80%BC)
已知a、b、c、是三个不等于零的有理数.1、若a+b+c=1,a^2+b^2+c^2=1,求1/a+1/b+1/c2、若a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3,求a+b+c的值
已知a、b、c、是三个不等于零的有理数.1、若a+b+c=1,a^2+b^2+c^2=1,求1/a+1/b+1/c
2、若a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3,求a+b+c的值
已知a、b、c、是三个不等于零的有理数.1、若a+b+c=1,a^2+b^2+c^2=1,求1/a+1/b+1/c2、若a^2+b^2+c^2=1,a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3,求a+b+c的值
1、a+b+c=1,
则(a+b+c)^2=1^2
a^2+b^2+c^2+2ab+2bc+2ac=1
∵a^2+b^2+c^2=1
∴a^2+b^2+c^2+2ab+2bc+2ac=1+2ab+2bc+2ac=1
∴2ab+2bc+2ac=0即ab+bc+ac=0
1/a+1/b+1/c=(bc+ac+ab)/abc=0/abc=0
2、题目抄错了吧,无解!
把a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3改成a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3有解
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3等式两侧都乘以abc,并去括号可得到
a^2×c+a^2×b+b^2×c+b^2×a+c^2×b+c^2×a=-3abc
移项得
a^2×c+c^2×a+abc+a^2×b+b^2×a+abc+b^2×c+c^2×b+abc=0
ac(a+b+c)+ab(a+b+c)+bc(a+b+c)=0
(a+b+c)(ac+ab+bc)=0
由题意可知a+b+c≠0,因此ac+ab+bc=0
a^2+b^2+c^2=1
a^2+b^2+c^2+2(ac+ab+bc)=1
即(a+b+c)^2=1
因此a+b+c=1或a+b+c=-1
发的萨芬的撒
题目抄错了吧,无解!
把a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3改成a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3有解
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3等式两侧都乘以abc,并去括号可得到
a^2×c+a^2×b+b^2×c+b^2×a+c^2×b+c^2×a=-3abc...
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题目抄错了吧,无解!
把a(1/b+1/c)+b(1/a+1/c)+(1/a+1/b)= -3改成a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3有解
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)= -3等式两侧都乘以abc,并去括号可得到
a^2×c+a^2×b+b^2×c+b^2×a+c^2×b+c^2×a=-3abc
移项得
a^2×c+c^2×a+abc+a^2×b+b^2×a+abc+b^2×c+c^2×b+abc=0
ac(a+b+c)+ab(a+b+c)+bc(a+b+c)=0
(a+b+c)(ac+ab+bc)=0
由题意可知a+b+c≠0,因此ac+ab+bc=0
a^2+b^2+c^2=1
a^2+b^2+c^2+2(ac+ab+bc)=1
即(a+b+c)^2=1
因此a+b+c=1或a+b+c=-1
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