用3种方法(破十法、连减法、想加算减法)完成16-8 13-7两个算式? 理工学科是什么
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完成16-8 13-7两个算式?解题思路:将减数与被减数个位对齐,在分别与对应计数单位上的数相减,不够减的需向高位借1,依次计算可以得出结果,减数小于被减数将两数调换相减最后结果加个负号;需要注意借位之后,高位的数需要减一;
解题过程:
16-8
步骤一:16-8=8 借位
步骤二:0-0=0
步骤三:计算结果为8
13-7
步骤一:13-7=6 借位
步骤二:0-0=0
步骤三:计算结果为6
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用三种方法进行这些算式的计算,那第1个16-8,破10法就是16=10+6拆分进行计算,连减法就是16直接减8。
绛旓細姝ラ涓:16-8=8 鍊熶綅 姝ラ浜:0-0=0 姝ラ涓:璁$畻缁撴灉涓8 13-7 姝ラ涓:13-7=6 鍊熶綅 姝ラ浜:0-0=0 姝ラ涓:璁$畻缁撴灉涓6 瀛樼枒璇疯拷闂紝婊℃剰璇烽噰绾
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