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展开全部Y = AB‘ + A‘C + C‘D‘ + D= AB‘ + A‘C + C‘D‘ + D + C‘D= AB‘ + A‘C + C‘ + D= AB‘ + A‘C + C‘ + A'C' + D= AB‘ + A‘ + C‘ + D= AB‘ + A‘ + A'B' + C‘ + D= A‘ + B' + C‘ + D
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选择擅长的领域继续答题？
{@each tagList as item}
${item.tagName}
{@/each}
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展开全部 Y=A'B+A'BC'D+BD'',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign),e.getAttribute("jubao"))},getILeft:function(t,e){return t.left+e.offsetWidth/2-e.tip.offsetWidth/2},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#href\}\}/g,e).replace(/\{\{#jubao\}\}/g,n)}},baobiao:{triangularSign:"data-baobiao",tpl:'{{#baobiao_text}}',getTip:function(t,e){return t.renderTip(e.getAttribute(t.triangularSign))},getILeft:function(t,e){return t.left-21},getSHtml:function(t,e,n){return t.tpl.replace(/\{\{#baobiao_text\}\}/g,e)}}};function l(t){return this.type=t.type
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我不会这两个题的化简，求解答，谢谢不会的就别写回答了，瞎整的东西有用吗...
我不会这两个题的化简，求解答，谢谢不会的就别写回答了，瞎整的东西有用吗
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选择擅长的领域继续答题？
{@each tagList as item}
${item.tagName}
{@/each}
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提交成功是否继续回答问题？
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①解A异或B可以表示成：AB非+A非B ，用这样的形式去掉原式中的“异或”“同或”。再根据摩根定律和逻辑代数的方法一般就可以化简成 最简与或式。如果遇到原式十分复杂的情况，可以用“卡诺图”来化简，其步骤一般是：（1）将逻辑函数写成最小项表达式（2）按最小项表达式填卡诺图，式中包含了的最小项其相应位置填1，其余填0（3）合并最小项，将相邻的“1”圈成一组，每组含2^n个方格，每个包围圈对应一个新的乘积项（即一个最简与或项）。（4）将所有包围圈对应的乘积项相加。得到最简与或表达式。②解几进制的计数器就是看该计数器计数计了多少个数然后返回初值。比如一个计数器从1开始计数，计到11的时候其下一个状态又变成1了，那么就是一个11进制的计数器。③不好意思，这个我不确定哈，我估计是为了匹配逻辑门的输出电平。可以看成是一个上拉电阻。④解首先，逻辑电路的每个状态对应一个圈，圈内的数字是来表示该状态。圈之间一般有箭头，箭头表示状态的转换，即从一个状态转换到另一个状态。箭头上面的数字表示状态转换需要的条件。具体表示的是那个量，还要根据画出状态图的状态表。表头上肯定会写出的。最后题，解CP表示Clock Pulse 是工作时钟，图示的JK触发器根据JK触发器的逻辑表达式，当J=K=1时，输出的下一个状态 = 目前状态 取非。即Qn+1=Qn非。数字电子技术基础—试题—解答-电子技术化简与或式...三、逻辑函数化简（每题5分，共10分）1、用代数法化简为最简与或式Y=A+1、Y=A+B2、用卡诺图法化简为最简或与式Y=+C+AD，约束条件：AC+ACD+AB=02、用卡诺图圈0的方法可得：Y=（+D）（A+）（+）四、分析下列电路。（每题6分，共12分）1、写出如图4所示电路的真值表及最简逻辑表达式。图41、该电路为三变量判一致电路，当三个变量都相同时输出为1，否则输出为0。2、写出如图5所示电路的最简逻辑表达式。2、B=1，Y=A，B=0Y呈高阻态。五、判断如图6所示电路的逻辑功能。若已知uB=-20V，设二极管