小于等于

使用排序不等式证：由幂指数a^a易知a,b,c>0，不妨假设a>b>c>0则a^a*b^b*c^c≥(abc)^[(a+b+c)/3]<=>ln(a^a*b^b*c^c)≥[(a+b+c)/3]lnabc对不等式左右取自然对数<=>alna+blnb+clnc≥[(a+b+c)/3]lnabc<=>3(alna+blnb+clnc)≥(a+b+c)lnabc即证3(alna+blnb+clnc)≥(a+b+c)lnabc由排序不等式3(alna+blnb+clnc)≥alna+blnb+clnc顺序和+blna+clnb+alnc乱序和1+clna+alnb+blnc乱序和2=(a+b+c)lna+(a+b+c)lnb+(a+b+c)lnc=(a+b+c)(lna+lnb+lnc)=(a+b+c)lnabc即3(alna+blnb+clnc)≥(a+b+c)lnabc当且仅当a=b=c时取等易证得当a,b,c取其他关系时同有上述式子成立故a^a*b^b*c^c≥(abc)^[(a+b+c)/3]得证当且仅当a=b=c时取等*含幂代数式通常对(不)等式两侧取自然对数利于变形放缩**对形如3(a1b1+a2b2+a3b3)一类式子需要灵活处理，向目标式拼凑转化

构造函数f(x)=xlnx，f"(x)=lnx+1，f""(x)=1/x>0所以f(x)下凸，由下凸函数的性质(即Jensen不等式)得f[(a+b+c)/3]≤[f(a)+f(b)+f(c)]/3即 [(a+b+c)/3]u2022ln[(a+b+c)/3]≤(alna+blnb+clnc)/3而 (abc)^(1/3)≤(a+b+c)/3，所以[(a+b+c)/3]u2022ln[(abc)^(1/3)]≤(alna+blnb+clnc)/3所以 ln{(abc)^[(a+b+c)/3)]}≤ln[(a^a)(b^b)(c^c)]从而 (abc)^[(a+b+c)/3)]≤(a^a)(b^b)(c^c)

你是问两个电对的电极电势的差值吧？0.36V即可

指的是开启电压，最小0.6V，最大1.2V。就是这个型号的管子的开启电压（GS之间加上一定的电压刚刚好使DS之间开始导通，这个电压就是开启电压）从0.6到1.2V之间，每个管子都有所不同，这是制造时的离散性造成的。

这是理论解决和实验都验证的，由于光波其实是不连续的，是一段段的脉冲，看我对你的另外的那个问题的解答补答：首先惠更斯原理是一种比较简单易懂的解释，它在经典光学里面有很重要的作用，在以前人们用它来解释一些物理现象很成功。当然了，物理发展到现在，最好的解释就是要以量子理论为基础的。你问到：障碍物比波长长时，障碍物后面的场的光强是为什么几乎平均的，你想象一下，波长是550纳米的光波经过障碍物时，我强调过;光波其实是不连续的，是一段段的脉冲，那么这时你就要想象一下两条只有550纳米长的振动的弦，一条是从上沿传出，一条是从下沿传出，如果障碍物的尺度比波长（550纳米）大，那么那两条弦经过障碍物后，你要在障碍物后面选一个点，看它们能不能相遇或那两根那么短的弦出现一小段重合？请注意，它们是同时到达上沿和下沿的，然后它们传播速度是一样的，这会你明白了吧，产生干涉的条件是要求干涉的那两条弦的振动频率是极其接近的，如果障碍物的尺度比波长（550纳米）大，那么那两条弦经过以后没有叠加的可能，这样就没有衍射的现象了，注意衍射和干涉本质上是一样的。如果障碍物比波长大，惠更斯原理同样适用。我为了解释的更容易明白，有一些地方是不严格的，比如让你把光脉冲想象成一条振动弦了

beqbq850

利于雨水排水

rab等于ra说明：R（A）是矩阵A的秩，R（AB），AB是A与B两个矩阵的积，也是一个矩阵，R（AB）就是这个积矩阵的秩。秩是线性代数术语。在线性代数中，一个矩阵的秩是其非零子式的最高阶数，一个向量组的秩则是其最大无关组所含的向量个数。在解析几何中，矩阵的秩可用来判断空间中两直线、两平面及直线和平面之间的关系。

c级点验钞机的漏辨率应小于等于0.03%。漏辨率是指未辨出的假币张数与实际清点假币张数的比率，简而言之就是，漏辨率过高，不是说完全验不出假币，而是验出率不够。国标对于C级验钞仪的漏辨率规定是不超过0.03%。点验钞机是一种自动清点钞票数目的机电一体化装置，一般带有伪钞识别功能，集计数和辨伪钞票的机器。由于现金流通规模庞大，银行出纳柜台现金处理工作繁重，点钞机已成为不可缺少的设备。随着印刷技术、复印技术和电子扫描技术的发展，伪钞制造水平越来越高，必须不断提高点钞机的辨伪性能。点验钞机的工作原理——验钞机的核心部件就是传感器，每一类型的传感器对应着一种鉴别技术，比如荧光鉴别、磁性鉴别等。传感器类型越多，意味着鉴别技术越全面，同时也代表着成本会更高。点验钞机辨伪原理包括荧光检测、磁性检测、红外穿透、激光检测、多光谱检测、分析检测等，其中磁性检测的工作原理是利用大面额真钞（20、50、100元）的某些部位是用磁性油墨印刷，通过一组磁头对运动钞票的磁性进行检测。

记忆儿歌：小小两条鱼，长得真稀奇。长着大嘴巴，名叫大小于。嘴巴向左边，名字叫大于。嘴巴向右边，名字叫小于。记忆口诀：大于号，小于号，两个兄弟一起到。尖头在前是小于，开口在前是大于。两个数字中间站，谁大冲谁开口笑。总结：大于号和小于号都是一头尖，一头张开大口，像一只鳄鱼张开大嘴巴。不一样的是，前面开口，后面尖尖的是大于号，它表示前面的物品或数字比后面的要多、要大;前面尖尖，后面开口的是小于号，它表示前面的物品或数字比后面的要少，要小。

至少，大于等于

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这里不受重用

看来你对变上限积分求导概念没有建立，一般地，对形如：F(x)=∫(0,g(x))f(t)φ(t)dt，若f(x)，φ(x)在g(x)的值域范围内连续，可导，那么：F"(x)=f[g(x)]·φ[g(x)]·g(x)该定理可用积分和导数的定义来证明，这里略根据题意，令h(t)=f.

级配是集料各级粒径颗粒的分配情况,可通过筛析试验确定. 建筑中的级配一般是指：砂,石的级配.在土木工程中,粒径大于5mm的骨料为粗骨料,又称为“石子”；粒径小于5mm的骨料为细骨料,又称为“砂”.我们可以通过筛分析,计算砂子的大小搭配状况,判断砂子的级配和细度模数.

受腊钢筋直径（mm）

因[(a+b+c)/3]^3 =(1/27)(a+b+c)(a+b+c)^2 =(1/27)(a+b+c)[(a^2+b^2+c^2)+2(ab+bc+ca)] 而由基本不等式或排序不等式有 2(ab+bc+ca)≤2(a^2+b^2+c^2)（a=b=c时取等） 则[(a+b+c)/3]^3 ≤(1/9)(a+b+c))(a^2+b^2+c^2)（a=b=c时取等） =(1/9)[(a^3+b^3+c^3)+(ab^2+bc^2+ca^2)+(a^2b+b^2c+c^2a)] 又由排序不等式有 ab^2+bc^2+ca^2≤a^3+b^3+c^3（a=b=c时取等） a^2b+b^2c+c^2a≤a^3+b^3+c^3（a=b=c时取等） 所以[(a+b+c)/3]^3≤(a^3+b^3+c^3)/3（a=b=c时取等） 即原命题成立 说明： 令a≤b≤c 则a^2≤b^2≤c^2 依据排序不等式,以上两组数的同序积和不小于乱序积和 这里a^3+b^3+c^3为同序积和 而ab^2+bc^2+ca^2、a^2b+b^2c+c^2a都是乱序积和