1.设a+b+c=0,求(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)的值2.设a,b,c满足1/a+1/b+1/c=1/(a+b+c),求证:1/(a^2n-1)+1/(b^2n-1)+1/(c^2n-1)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]
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![1.设a+b+c=0,求(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)的值2.设a,b,c满足1/a+1/b+1/c=1/(a+b+c),求证:1/(a^2n-1)+1/(b^2n-1)+1/(c^2n-1)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]](/uploads/image/z/10248652-28-2.jpg?t=1.%E8%AE%BEa%2Bb%2Bc%3D0%2C%E6%B1%82%28a%5E2%2F2a%5E2%2Bbc%29%2B%28b%5E2%2F2b%5E2%2Bac%29%2B%28c%5E2%2F2c%5E2%2Bab%29%E7%9A%84%E5%80%BC2.%E8%AE%BEa%2Cb%2Cc%E6%BB%A1%E8%B6%B31%2Fa%2B1%2Fb%2B1%2Fc%3D1%2F%28a%2Bb%2Bc%29%2C%E6%B1%82%E8%AF%81%EF%BC%9A1%2F%EF%BC%88a%5E2n-1%29%2B1%2F%28b%5E2n-1%EF%BC%89%2B1%2F%28c%5E2n-1%29%3D1%2F%5B%28a%5E2n-1%29%2B%28b%5E2n-1%29%2B%28c%5E2n-1%29%5D)
1.设a+b+c=0,求(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)的值2.设a,b,c满足1/a+1/b+1/c=1/(a+b+c),求证:1/(a^2n-1)+1/(b^2n-1)+1/(c^2n-1)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]
1.设a+b+c=0,求(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)的值
2.设a,b,c满足1/a+1/b+1/c=1/(a+b+c),求证:1/(a^2n-1)+1/(b^2n-1)+1/(c^2n-1)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]
1.设a+b+c=0,求(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)的值2.设a,b,c满足1/a+1/b+1/c=1/(a+b+c),求证:1/(a^2n-1)+1/(b^2n-1)+1/(c^2n-1)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]
1.(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)=3/2+ac+bc+ab
因为a+b+c=0,所以(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=0,所以ab+bc+ac=0
所以有(a^2/2a^2+bc)+(b^2/2b^2+ac)+(c^2/2c^2+ab)=3/2+0=3/2
2.令A=a^2n-1,B=b^2n-1,C=c^2n-1
根据题意有
1/A+1/B+1/C=1/(A+B+C)=1/[(a^2n-1)+(b^2n-1)+(c^2n-1)]
1。把题转化为a+b+c=0的形式就可以了。