﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算请顺便讲讲道理吧!

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/17 20:52:54
﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算请顺便讲讲道理吧!
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﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算请顺便讲讲道理吧!
﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算
请顺便讲讲道理吧!

﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算请顺便讲讲道理吧!
设 0.112+0.123=a
原式=﹙1+a﹚×﹙a+0.134﹚-﹙1+a+0.134﹚×a
=a+a²+0.134a+0.134 - (a+a²+0.134a)
=0.134

﹙1+0.112+0.123﹚×﹙0.112+0.123+0.134﹚-﹙1+0.112+0.123+0.134﹚×﹙0.112+0.123﹚简便计算请顺便讲讲道理吧! ﹙1-二分之一﹚×﹙1-三分之一﹚×﹙1-四分之一﹚×...×﹚1-九十九分之一﹚×﹙1 [-0.5²+﹙-1/2﹚²-|-2²-4|+﹙2 又1/4﹚²×26/27]÷﹙0.1﹚² ∫﹙lnx-1﹚/﹙x²﹚dx 分式加减法计算 ﹙1﹚/﹙x+1﹚+﹙2﹚/﹙x+1﹚ 分式加减法计算 ﹙1﹚/﹙x+1﹚+﹙2﹚/﹙x+1﹚ 求值:﹙1﹚arc sin﹙sin1﹚ ﹙2﹚arc cos﹙cosπ/12﹚ ﹙x-½﹚﹙½+x﹚﹣﹙x﹣1﹚﹙x+¼﹚ ﹙﹣2﹚×﹙﹣7﹚×﹙﹢5﹚×[﹣7分之1] 设定义在R上的函数f﹙x﹚满足对于任意x,y属于R都有f﹙x+y﹚=f﹙x﹚﹢f﹙y﹚成立,且f﹙1﹚=-2,当x>0时,f﹙x﹚﹤0.1﹚判断f﹙x﹚的奇偶性,并加以证明2﹚试问:当-3≤x≤3时,f﹙x﹚是否有最值?有, 用简便方法计算:1+﹙-2﹚+3+﹙-4﹚+5+﹙-6﹚+7+﹙-8﹚+...+99+﹙-100﹚= ﹙2x+5﹚2+﹙x-1﹚﹙3-x﹚-﹙x+4﹚﹙2-3x﹚ ﹣2×﹙0.1﹚的立方×﹙﹣10﹚的平方+﹙﹣0.8﹚ -3²+﹙-2﹚³-﹙0.1﹚²×﹙-10﹚³ ﹙-36 12/29﹚÷12 计算(㏒10﹙8)+㏒10﹙1000﹚﹚×㏒10﹙5﹚+﹙㏒10﹙2√3﹚2+㏒10﹙1/6﹚+㏒10﹙0.06﹚ 计算(㏒10﹙8)+㏒10﹙1000﹚﹚×㏒10﹙5﹚+﹙㏒10﹙2√3﹚2+㏒10﹙1/6﹚+㏒10﹙0.06﹚ 计算:要写清楚计算过程!①:﹙﹣5﹚-﹙﹣5﹚×5分之1×﹙﹣4﹚ ②:﹙﹣1﹚×﹙﹣7﹚+6×﹙﹣1﹚×2分之1 ③:﹙﹣3﹚×﹙﹣7﹚-3×﹙﹣6﹚ ④:1-﹙﹣1﹚×﹙﹣1﹚-﹙﹣1﹚×0×﹙﹣1﹚ 急 ﹙1+0.23+0.34﹚×﹙0.23+0.34+0.56)-﹙1+0.23+0.34+0.56﹚×﹙0.23+0.34)