Question

（20又9/4+12.25-31又30/1)除299+9（13又9/1-12.25）*1又31/5（过程详细点）

Answer 1

(20又9/4+12.25-31又30/1)÷299+9(13又9/1-12.25)×1又31/5

=(9/184+4/49-30/931)×299/1+9(9/118-4/49)×31/36

=[(9×20)/(184×20)+(4×45)/(49×45)-(30×6)/(931×6)]×299/1+9[(9×4)/(118×4)-(4×9)/(49×9)]×31/36

=(180/3680+180/2205-180/5586)×299/1+9(36/472-36/441)×31/36

=180/(3680+2205-5586)×299/1+9×36/(472-441)×31/36

=180/299×299/1+9×36/31×31/36

=180/1+9

=9又180/1

最后结果等于9又180分之1