题目内容
计算:(12.345+23.451+34.512+45.123+51.234)÷(1+2+3+4+5)= .
考点:四则混合运算中的巧算
专题:计算问题(巧算速算)
分析:把第一个括号内的每个数字进行拆分,例如12.345=1×10+2×1+3×0.1+4×0.01+5×0.001,…,运用乘法分配律计算,最后再算除法,解决问题.
解答:
解:(12.345+23.451+34.512+45.123+51.234)÷(1+2+3+4+5)
=[(1×10+2×1+3×0.1+4×0.01+5×0.001)+(2×10+3×1+4×0.1+5×0.01+1×0.001)+(3×10+4×1+5×0.1+1×0.01+2×0.001)+(4×10+5×1+1×0.1+2×0.01+3×0.001)+(5×10+1×1+2×0.1+3×0.01+4×0.001)]÷(1+2+3+4+5)
=[(1+2+3+4+5)×10+(1+2+3+4+5)×1+(1+2+3+4+5)×0.1+(1+2+3+4+5)×0.01+(1+2+3+4+5)×0.001]÷(1+2+3+4+5)
=[(1+2+3+4+5)×(10+1+0.1+0.01+0.001)])÷(1+2+3+4+5)
=(1+2+3+4+5)×11.111÷(1+2+3+4+5)
=11.111
故答案为:11.111.
=[(1×10+2×1+3×0.1+4×0.01+5×0.001)+(2×10+3×1+4×0.1+5×0.01+1×0.001)+(3×10+4×1+5×0.1+1×0.01+2×0.001)+(4×10+5×1+1×0.1+2×0.01+3×0.001)+(5×10+1×1+2×0.1+3×0.01+4×0.001)]÷(1+2+3+4+5)
=[(1+2+3+4+5)×10+(1+2+3+4+5)×1+(1+2+3+4+5)×0.1+(1+2+3+4+5)×0.01+(1+2+3+4+5)×0.001]÷(1+2+3+4+5)
=[(1+2+3+4+5)×(10+1+0.1+0.01+0.001)])÷(1+2+3+4+5)
=(1+2+3+4+5)×11.111÷(1+2+3+4+5)
=11.111
故答案为:11.111.
点评:对于类似题目,也可直接运用公式计算.12.345+23.451+34.512+45.123+51.234=11.111×(1+2+3+4+5),然后除以(1+2+3+4+5),解决问题.
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