递等式计算.能简便要简便.

题目

递等式计算.能简便要简便.
7/31×13+13/31×20
2/11×（11＋44）＋3/5÷1/25
（6又1/2-3又3/4）÷(13＋11又1/5)
3.3×[38＋3又1/5÷（1又3/4-0.15）]

答案

解1：
7/31×13+13/31×20
=7×13/31+13×20/31
=(13/31)(7+20)
=(13/31)×27
=13×27/31
=351/31
解2：
(2/11)×(11+44)+(3/5)÷1/25
=(2/11)×55+(3/5)×25
=2×5+3×5
=10+15
=25
解3：
(6又1/2-3又3/4)÷(13＋11又1/5)
=(13/2-15/4)÷(13+56/5)
=(26/4-15/4)÷(65/5+56/5)
=(11/4)÷(121/5)
=(11/4)×(5/121)
=(1/4)×(5/11)
=5/44
解4：
3.3×[38+3又1/5÷(1又3/4-0.15)]
=(33/10)×[38+(16/5)÷(7/4-3/20)]
=(33/10)×[38+(16/5)÷(35/20-3/20)]
=(33/10)×[38+(16/5)÷(8/5)]
=(33/10)×[38+(16/5)×(5/8)]
=(33/10)×(38+2)
=(33/10)×40
=33×4
=132