问题
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:192×192×192-171×171×171=()。A.1905258 B.2066755 C.2077677D.3217509
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已知某主机的IP地址为:192.168.100.200,子网掩码为:255.255.255.192,则该主机所在的网络地址:(27
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A.192.168.100.224B.192.168.100.255C.192.168.100.248D.192.168.100.216
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将192.168.100.0/24和192.168.106.0/24地址聚合应该选择那种表达方式()。A.192.168.0.0/24B.192.1
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以下()地址是192.168.15.19/28子网内的主机可用地址。A.192.168.15.16B.192.168.15.29C.192.168
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已知某主机的IP地址为:192.168.100.200 子网掩码为:255.255.255.192 可推导出()