解：原式可以采用分步除法来解答
(3\4+2\3+1\6)÷1\12
=3/4÷1/12+2/3÷1/12+1/6÷1/12
=3/4*12+2/3*12+1/6*12
=9+8+2
=19
(3/4+2/3+1/6)÷1/12
=(3/4+2/3+1/6)x12
=3/4x12+2/3x12+1/6x12
=9+8+2
=19
谢谢，请采纳
（3/4+2/3+1/6）÷1/12
＝（3/4+2/3+1/6）×12
＝3/4×12+2/3×12+1/6×12
＝9+8+2
＝19
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