20# reny
那肯定麻烦了噻！随便试试看：
\[y=\frac{a_1} {x-b_1}+\frac{a_2}{x-b_2}+\frac1 {b_3-x}\],
\[y=\frac{1}{\dfrac x{a_1}-\dfrac{b_1}{a_1}}+\frac{1}{\dfrac x{a_2}-\dfrac{b_2}{a_2}}+\frac1{b_3-x}\],
令$x=a_1a_2t$,则
\[y=\frac{1}{a_2t-\dfrac{b_1}{a_1}}+\frac{1}{a_1t-\dfrac{b_1}{a_1}}+\frac1 {b_3-a_1a_2t}\],
作代换后，
\[y=\frac{1}{a_2t-a}+\frac{1}{a_1t-b}+\frac1{c-a_1a_2t}\],
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发表于 2013-3-7 13:46