已知abc≠0,a+b+c=1,a^2+b^2+c^2=1,求(2b+2c)/a+(2a+2b)/b+(2a+2b)/c的值(2b+2c)/a+(2a+2c)/b+(2a+2b)/c=2(b+c)/a+2(a+c)/b+2(a+b)/c=2(1-a)/a+2(1-b)/b+2(1-c)/c=2(1/a+1/b+1/c)-6=2[(bc+ac+ab)/(abc)]-6,因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,所以1=1+2(a

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已知abc≠0,a+b+c=1,a^2+b^2+c^2=1,求(2b+2c)/a+(2a+2b)/b+(2a+2b)/c的值(2b+2c)/a+(2a+2c)/b+(2a+2b)/c=2(b+c)/a+2(a+c)/b+2(a+b)/c=2(1-a)/a+2(1-b)/b+2(1-c)/c=2(1/a+1/b+1/c)-6=2[(bc+ac+ab)/(abc)]-6,因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,所以1=1+2(a
已知abc≠0,a+b+c=1,a^2+b^2+c^2=1,求(2b+2c)/a+(2a+2b)/b+(2a+2b)/c的值
(2b+2c)/a+(2a+2c)/b+(2a+2b)/c
=2(b+c)/a+2(a+c)/b+2(a+b)/c
=2(1-a)/a+2(1-b)/b+2(1-c)/c
=2(1/a+1/b+1/c)-6
=2[(bc+ac+ab)/(abc)]-6,
因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,
所以1=1+2(ab+ac+bc),
所以ab+bc+ac=0,
所以原式=2[(bc+ac+ab)/(abc)]-6
=2*0-6
=-6.
以上的解我看不懂,请高人予以解析,或者写个更清晰易懂的过程给我,
=2(1-a)/a+2(1-b)/b+2(1-c)/c
=2(1/a+1/b+1/c)-6
我看不懂这步.

已知abc≠0,a+b+c=1,a^2+b^2+c^2=1,求(2b+2c)/a+(2a+2b)/b+(2a+2b)/c的值(2b+2c)/a+(2a+2c)/b+(2a+2b)/c=2(b+c)/a+2(a+c)/b+2(a+b)/c=2(1-a)/a+2(1-b)/b+2(1-c)/c=2(1/a+1/b+1/c)-6=2[(bc+ac+ab)/(abc)]-6,因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,所以1=1+2(a
(2b+2c)/a+(2a+2c)/b+(2a+2b)/c
=2(b+c)/a+2(a+c)/b+2(a+b)/c
=2(1-a)/a+2(1-b)/b+2(1-c)/c {因为a+b+c=1,所以b+c=1-a,其他以此类推}
=2(1/a+1/b+1/c)-6
=2[(bc+ac+ab)/(abc)]-6,
因为(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc,
所以1=1+2(ab+ac+bc),
所以ab+bc+ac=0,
所以原式=2[(bc+ac+ab)/(abc)]-6
=2*0-6
=-6.