• 2022-07-02
    试证: [tex=8.5x2.571]N9BpSvrTmt3CxRPEGcvCN/LLtvjn6bmtuaGFf/vcTHrVf2qqV2/VjKrjQAlaJr1hW751eNPkGhn0mkCDreWaaF9KbqcdZFvRKCnOHj/sX8I=[/tex].提示:等式右边满足 [tex=1.857x1.357]4zsKoWClgoMa84syvwy2EQ==[/tex] 定义,并注意 [tex=7.357x2.857]KYDD1VVtyjGfkmVyl5f5vlUbcFycfF1et5K/2QZLu0CFyBHojKneu0mtYKkfX5K6GnXAYXnI55gTJHEUPYq8KQ==[/tex]
  • 证明:当 [tex=1.857x1.0]bOlCq/PHWhsSVMaVf7Obdg==[/tex] 时,考虑到[tex=14.643x2.857]+BGXk3A1bsNMt1R+djuo9Oivv9r8zjtUA8Sp1P82kFc4QINg0ksO+y2TEytz+r51iTgesuR9cBdqPV/En4NdAvN+NoULf6Q6h1rgHo3V92UIOxAP8B+391SB+aeeAsV/o60AiynoIwIgyf2pmxOSbRrPvrJr3QDu1/uHweVBCkTTr5WTGartkSJFOkzh2Gsw[/tex]由当 [tex=2.429x1.214]J/cCyv8nBbLutvPfBXF+Yw==[/tex] 时,有[tex=9.286x2.571]0HWBscUXKnK8yOYhpSVA9a7q3ke/+ZcFYDZzhgTWBubQ07cf4TGO6n/HdyYEkzO3EJ2Qrx/zeZsikY33Wuqlz1Oiad+UbVXGnhWUrl2Dtks=[/tex]根据夹逼准则可得[tex=7.214x2.571]+BGXk3A1bsNMt1R+djuo9E+kMC/neJ750PCA3Pqb0GDTGAdegtRMgjpebEDsa6nr6j6cUbP4QBkNpky4qlTsAkbBfhYwiJdyYgF7Mx5PCWw=[/tex]即[tex=13.286x2.929]+BGXk3A1bsNMt1R+djuo9E+kMC/neJ750PCA3Pqb0GDTGAdegtRMgjpebEDsa6nr6j6cUbP4QBkNpky4qlTsAk7n3NrCgxT7hVSMMetIpJBJPr1HBbXjVKKjKRo2b2pK8VgFWwO0KaGbidUH9BsLfzS1pcoxMCp36yzhzZB5yDahLXbfaJg6oLeR2VzD3uzl[/tex]又有[tex=9.5x9.714]KYDD1VVtyjGfkmVyl5f5vlUbcFycfF1et5K/2QZLu0CmDzEQSNjcFkhxdbIY8/8LZFD7fIUv/0Gy+b+bs93xZFDIKZAdJrl3XaJOQcb3QTUN9R/hMOrYcrVz7zH3b8sQLaLHL399rDw4SxrRV5tGBJ2bodAjKJALvQf5w7lKMbUSS08iFhGb0FIqRUYZCHeLRPis7t6N+cAJUV0QpvWFGdCX4SNXKTls9Ky0MYhbaO4FeDVxwNRTzAOEn/iKWwkpmFBOVETWfLR/cfKHBG/hkCXvPyDAKhEl51IB92W5Oac=[/tex]所以[tex=11.0x2.857]KYDD1VVtyjGfkmVyl5f5vn1MP5qjRvG0ly3R5301mwlndehAE/23DnEc/NBvI/fAIrNL0u5nMBLSfIT5CyDh8/EBE9BvwS4Ltf0oITNCMWh7r2hpghNWS7P7c5GKDL7h[/tex]即 [tex=8.5x2.571]N9BpSvrTmt3CxRPEGcvCN/LLtvjn6bmtuaGFf/vcTHrVf2qqV2/VjKrjQAlaJr1hW751eNPkGhn0mkCDreWaaF9KbqcdZFvRKCnOHj/sX8I=[/tex]

    内容

    • 0

      ‌下面说法错误的是( )。‌‌知识点:列表推导式‌ A: dict([(x, x**2) for x in range(6)]) 创建的字典是{0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25} B: [[x*3+y for y in range(1,4)] for x in range(3)] 创建的是二维列表 [[1, 2, 3], [4, 5, 6], [7, 8, 9]] C: number = [-2, 4, 6, -5]string = 'ab'z = [(i, j) if i>0 else (-i, j) for i in number for j in string]这段代码创建的列表为[(2, 'a'), (2, 'b'), (4, 'a'), (4, 'b'), (6, 'a'), (6, 'b'), (5, 'a'), (5, 'b')] D: ' '.join([i for i in range(1,11)])的运算结果为字符串'1 2 3 4 5 6 7 8 9 10'

    • 1

      当执行下面的语句定义一维数组a后,此数组的所有元素为 ( ) inta[10]; A.a[1],a[2],a[3],a[4],a[5],a[6],a[8],a[9],a[10],a[10] B.a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9] C.a[0],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9],a[10] D.a[1],a[2],a [3],a[4],a [5],a [9],a [7],a [8],a [9],a [10],a [11]

    • 2

      下面语句的输出结果是?range(len('HelloWorld')) A: [1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11] B: 11 C: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] D: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]

    • 3

      for (int i = 1; i <= 10; i++){ if (i % 5 != 0) continue; Console.WriteLine("{0}", i); }该段程序执行后输出_____。 A: 5 10 B: 1 2 3 4 6 7 8 9 C: 5 D: 1 2 3 4 5 6 7 8 9 10

    • 4

      输出九九乘法表。 1 2 3 4 5 6 7 8 9 --------------------------------------------------------------------- 1*1=1 2*1=2 2*2=4 3*1=3 3*2=6 3*3=9 4*1=4 4*2=8 4*3=12 4*4=16 5*1=5 5*2=10 5*3=15 5*4=20 5*5=25 6*1=6 6*2=12 6*3=18 6*4=24 6*5=30 6*6=36 7*1=7 7*2=14 7*3=21 7*4=28 7*5=35 7*6=42 7*7=49 8*1=8 8*2=16 8*3=24 8*4=32 8*5=40 8*6=48 8*7=56 8*8=64 9*1=9 9*2=18 9*3=27 9*4=36 9*5=45 9*6=54 9*7=63 9*8=72 9*9=81