一条分式数学题,希望能有详细的解答.

已知a+b+c=0,试求 a^2/[2a^2+bc]+b^2/[2b^2+ac]+c^2/[2c^2+ab]的值

a+b+c=0=====>a+b=-c

a^3+b^3=(a+b)(a^2-ab+b^2)=-c[(a+b)^2-3ab]=-c(c^2-3ab)=3abc-c^3

a^2/[2a^2+bc]+b^2/[2b^2+ac]

=[a^2(2b^2+ac)+b^2(2a^2+bc)]/[(2a^2+bc)(2b^2+ac)]

=[4a^2b^2+c(a^3+b^3)]/[4a^2b^2+2c(a^3+b^3)+abc^2]

=[4a^2b^2+c(3abc-c^3)]/[4a^2b^2+2c(3abc-c^3)+abc^2]

=[4a^2b^2+3abc^2-c^4]/[4a^2b^2+6abc^2-2c^4+abc^2]

=[4a^2b^2+3abc^2-c^4]/[4a^2b^2+7abc^2-2c^4]

=[(4ab-c^2)(ab+c^2)]/[(4ab-c^2)(ab+2c^2)]

=(ab+c^2)/(ab+2c^2)

所以：a^2/[2a^2+bc]+b^2/[2b^2+ac]+c^2/[2c^2+ab]

=(ab+c^2)/(ab+2c^2)+c^2/(2c^2+ab)

=(ab+c^2+c^2)/(2c^2+ab)

=(2c^2+ab)/(2c^2+ab)

=1