题目内容
用二进制计算:
(1)101110(2)+1001101(2);
(2)1001001(2)-101110(2);
(3)1001001(2)×1001(2);
(4)1110101(2)÷1101(2);
(5)1001(2)×1110(2)÷10101(2).
(1)101110(2)+1001101(2);
(2)1001001(2)-101110(2);
(3)1001001(2)×1001(2);
(4)1110101(2)÷1101(2);
(5)1001(2)×1110(2)÷10101(2).
分析:根据二进制的加法,减法,乘法与除法运算法则计算即可求解.
解答:解:(1)101110(2)+1001101(2)=1111011(2);
(2)1001001(2)-101110(2)=11011(2);
(3)1001001(2)×1001(2)=1010010001(2);
(4)1110101(2)÷1101(2)=1001(2);
(5)1001(2)×1110(2)÷10101(2),
=1111110(2)÷10101(2),
=110(2).
(2)1001001(2)-101110(2)=11011(2);
(3)1001001(2)×1001(2)=1010010001(2);
(4)1110101(2)÷1101(2)=1001(2);
(5)1001(2)×1110(2)÷10101(2),
=1111110(2)÷10101(2),
=110(2).
点评:本题考查的知识点是二进制的加法,减法,乘法与除法运算,二进制加法与乘法的定义:0+0=0,0+1=1+0=1,1+1=10;0×0=0×1=1×0=0,1×1=1,是解答本题的关键.
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