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简便运算(1又5/99+3又5/33+9又5/11)/(1又1/99+3又1/33+9又1/11),
人气:169 ℃ 时间:2020-04-10 05:05:32
解答
(1又5/99+3又5/33+9又5/11)/(1又1/99+3又1/33+9又1/11)
=(13+5/99+5/33+5/11)/(13+1/99+1/33+1/11)
=1+4×(1/99+1/33+1/11)/(13+1/99+1/33+1/11)
=1+4×(1+3+9)/(13×99+1+3+9)
=1+4×13/(13×100)
=1+1/25
=1又1/25=1+4×(1/99+1/33+1/11)/(13+1/99+1/33+1/11)=1+4×(1+3+9)/(13×99+1+3+9)=1+4×13/(13×100) 这些过程是什么意思?(1又5/99+3又5/33+9又5/11)/(1又1/99+3又1/33+9又1/11)=(13+5/99+5/33+5/11)/(13+1/99+1/33+1/11)=(13+1/99+1/33+1/11+4/99+4/33+4/11)/(13+1/99+1/33+1/11)==(13+1/99+1/33+1/11)/(13+1/99+1/33+1/11)+(4/99+4/33+4/11)/(13+1/99+1/33+1/11)=1+4×(1/99+1/33+1/11)/(13+1/99+1/33+1/11)=1+4×(1+3+9)/(13×99+1+3+9)=1+4×13/(13×100)=1+1/25=1又1/251+4×(1+3+9)/(13×99+1+3+9)什么意思?=1+4×(1/99+1/33+1/11)/(13+1/99+1/33+1/11)被除数和除数同时×99=1+4×(1+3+9)/(13×99+1+3+9)=1+4×13/(13×100)=1+1/25=1又1/25
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