有些分数应用题含有不同的单位“1”,解这类题时,只要从已知条件中找出不变量,再寻突破口,问题就会迎刃而解。一、总量不变例1一个最简分数,分子加上3,约简得59;若分母加3,则成13。求原分数。分析与解:由两次分数变化都是加3,可知分子和分母的和虽然变化但仍然相等。因为59的分 Some fractional application questions contain different units “1”. When solving such questions, the problem can be easily solved by finding the invariants from the known conditions and looking for a breakthrough. First, the total invariant 1 A simplest fraction, plus 3 molecules, reduced to 59; if the denominator plus 3, then into 13. Find the original score. Analysis and reconciliation: changes by two scores are plus 3, we can see the numerator and denominator although the change but still equal. Because of the 59 points