(bc+b-c)/(b^2c^2)+(ca+c-a)/(c^2a^2)+(ab+a-b)/(a^2b^2)=a^2(bc+b-c)/(b^2c^2a^2)+b^2(ca+c-a)/(c^2a^2b^2)+c^2(ab+a-b-1)/(a^2b^2c^2)=(a^2bc+a^2b-a^2c+b^2ac+b^2c-b^2a+c^2ab+c^aa-c^2b)/(a^2b^2c^2) =[(a+b+c)a...
一条分式数学题,希望能有高手为我详细的解答.
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