一道数学难题,高人请进,好的赏200!

A=1.2x10^(-198) B=2.5x10^(-99) C=8x10^(-100)
A/BC
=1.2x10^(-198)/[2.5x10^(-99)x8x10^(-100)]
=12/20
=0.6

A=0.(0……0)12=12*10^(-201)

B=0.(0……0)25=25*10^(-102)

C=0.(0……0)8 =8*10^(-101)

那么A除以(B乘C)
＝A/(B*C)
=12*10^(-201+102+101)/(25*8)
=6

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+2