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主题目录

借条件分解分母

第 10 章01代数入门

例如：已知 a+b+c=1a+b+c=1，化简
T=1ab+c+1bc+a+1ca+b. T=\frac1{ab+c}+\frac1{bc+a}+\frac1{ca+b}.
用 c=1−a−bc=1-a-b 改写第一个分母：
ab+c=ab+1−a−b=(a−1)(b−1). ab+c=ab+1-a-b=(a-1)(b-1).
另外两个分母也能作相同变形。因此
T=1(a−1)(b−1)+1(b−1)(c−1)+1(c−1)(a−1)=(c−1)+(a−1)+(b−1)(a−1)(b−1)(c−1)=−2(a−1)(b−1)(c−1). \begin{aligned} T&=\frac1{(a-1)(b-1)}+\frac1{(b-1)(c-1)} +\frac1{(c-1)(a-1)}\\[4pt] &=\frac{(c-1)+(a-1)+(b-1)}{(a-1)(b-1)(c-1)}\\[4pt] &=\frac{-2}{(a-1)(b-1)(c-1)}. \end{aligned}
若需改写成基本对称式，由 a+b+c=1a+b+c=1，还有
(a−1)(b−1)(c−1)=abc−ab−bc−ca,T=−2abc−ab−bc−ca. (a-1)(b-1)(c-1)=abc-ab-bc-ca, \qquad T=\frac{-2}{abc-ab-bc-ca}.