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](/uploads/image/z/1009048-40-8.jpg?t=%E5%B7%B2%E7%9F%A5abc%E2%89%A00%2Ca%2Bb%2Bc%3D1%2Ca%5E2%2Bb%5E2%2Bc%5E2%3D1%2C%E6%B1%82%282b%2B2c%29%2Fa%2B%282a%2B2b%29%2Fb%2B%282a%2B2b%29%2Fc%E7%9A%84%E5%80%BC%282b%2B2c%29%2Fa%2B%282a%2B2c%29%2Fb%2B%282a%2B2b%29%2Fc%3D2%28b%2Bc%29%2Fa%2B2%28a%2Bc%29%2Fb%2B2%28a%2Bb%29%2Fc%3D2%281-a%29%2Fa%2B2%281-b%29%2Fb%2B2%281-c%29%2Fc%3D2%281%2Fa%2B1%2Fb%2B1%2Fc%29-6%3D2%5B%28bc%2Bac%2Bab%29%2F%28abc%29%5D-6%2C%E5%9B%A0%E4%B8%BA%28a%2Bb%2Bc%29%5E2%3Da%5E2%2Bb%5E2%2Bc%5E2%2B2ab%2B2ac%2B2bc%2C%E6%89%80%E4%BB%A51%3D1%2B2%28a)
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.