对于某些分式求值的题目，若能根据其结构特点，选择适当的方法进行运算，常可使运算简便．举例如下：一、整体代入法例１已知ａ＋ｂ＋ｃ＝３，ａｂ＋ｂｅ＋ｃａ＝２，求　　的值．解原式＝例２若　　　　　　则　　（１９９４年天津市中考试题）解　设　　　　则ａ＝２ｋ，ｂ For some topics of fractional evaluation, if you can select the appropriate method for calculation based on its structural characteristics, you can often make the operation simple. Examples are as follows: First, the whole substitution law 1 knows that a+b+c=3, ab+be+ca=2, the value to be found. Solution original = example 2 If (1994 Tianjin test questions) solution, a = 2k, b