已知abc是三个不等于0的有理数,且a+b+c=1 a^2+b^2+c^2=1求(1/a)+(1/b)+(1/c)
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已知abc是三个不等于0的有理数,且a+b+c=1 a^2+b^2+c^2=1求(1/a)+(1/b)+(1/c)
已知abc是三个不等于0的有理数,且a+b+c=1 a^2+b^2+c^2=1求(1/a)+(1/b)+(1/c)
已知abc是三个不等于0的有理数,且a+b+c=1 a^2+b^2+c^2=1求(1/a)+(1/b)+(1/c)
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac
a+b+c=1,a^2+b^2+c^2=1
所以ab+ac+bc=0
因为abc是三个不等于0的有理数,abc不等于0
将ab+ac+bc=0同除以abc,得出(1/a)+(1/b)+(1/c)=0
(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=1+2(ab+bc+ac)=1
ab+ac+bc=0
1/a+1/b+1/c=(bc+ac+ab)/abc=0
(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)
1^2=1+2(ab+ac+bc)
ab+ac+bc=0
1/a+1/b+1/c
=(bc+ac+ab)/abc
=0