已知a+b+c=0,1/a+1/b+1/c=﹣4,那么1/a²+1/b²+1/c²的值为
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可化为1/(a^2)+1/(b^2)+1/(c^2)
=(a^2b^2+b^2c^2+c^2a^2)/(a^2b^2c^2)
因为(1/a+1/b+1/c)^2=(a^2b^2+2ab^2c+b^2c^2+2bc^2a+c^2a^2+2ca^b)/(a^2b^2c^2)
比上面多了(2ab^2c+2bc^2a+2ca^b)/(a^2b^2c^2)=2(a+b+c)/(abc)〉〉
所以原式
=(1/a+1/b+1/c)^2-2(a+b+c)/(abc)
=16-0
=16
=(a^2b^2+b^2c^2+c^2a^2)/(a^2b^2c^2)
因为(1/a+1/b+1/c)^2=(a^2b^2+2ab^2c+b^2c^2+2bc^2a+c^2a^2+2ca^b)/(a^2b^2c^2)
比上面多了(2ab^2c+2bc^2a+2ca^b)/(a^2b^2c^2)=2(a+b+c)/(abc)〉〉
所以原式
=(1/a+1/b+1/c)^2-2(a+b+c)/(abc)
=16-0
=16
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(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16
参考资料: 知道
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(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2/bc
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16
(1/a+1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2(a+b+c)/abc
1/a^2+1/b^2+1/c^2=(1/a+1/b+1/c)^2-2(a+b+c)/abc
=(-4)^2-0
=16
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