设[tex=18.143x1.357]ES4bZhVD42u6rDfT+4RLbFADiX3Lf8M2IGFEPl2XyJ4VRmrY0PeXhFGOHpnggt6l+iW3nlHvOwMPqM/gn+piIKwbxCdMNTiZ8gifFjb0l7G2tJE8LTHarm2xS0To8zta0baWJJdKnSlw48lWN3eJh+U/YjutlgYvwceRYUN9/I8=[/tex]求[tex=12.643x1.5]MKahQX4IzmXQSpYZPPgJu+Zh7+QnAKBGe68pVWFAkiC87j9j2tIKdhpzwvFihfgaIxRN94JdI2fgdZS7l/EROA==[/tex]

设[tex=18.143x1.357]ES4bZhVD42u6rDfT+4RLbFADiX3Lf8M2IGFEPl2XyJ4VRmrY0PeXhFGOHpnggt6l+iW3nlHvOwMPqM/gn+piIKwbxCdMNTiZ8gifFjb0l7G2tJE8LTHarm2xS0To8zta0baWJJdKnSlw48lWN3eJh+U/YjutlgYvwceRYUN9/I8=[/tex]求[tex=12.643x1.5]MKahQX4IzmXQSpYZPPgJu+Zh7+QnAKBGe68pVWFAkiC87j9j2tIKdhpzwvFihfgaIxRN94JdI2fgdZS7l/EROA==[/tex]

  证明线性卷积服从交换律、结合律和分配律，即证明下面等式成立: [tex=18.143x1.357]HWRwPo+7bcM+XCdiqY/+Qumvs5KV7JzdWeI7xqQUeaCSsCrBNMwJuBZvC42muXFhImniG63/BZ2DHmnA0SulYyEj1odaSJfPetMdti/HeJs1qiky7mC9fAiM61/NTBUQ[/tex] 

定义下式为符号集大小为K的离散平稳信源的符号熵;[tex=18.143x1.357]ttWmeJ0n+ybVcSqG4ZbA3SJLN6ZNHGdcMsmUr2CGEO0POInyX+7KZrCmrIR8uUTfcpO7WB9SqzSqVQjlWSDplPS6zmW8qVGMjyBnFb1PgfQLSfwppb75603kCM4eBRt6[/tex]证明[tex=3.214x1.357]Ul8l0nTDCQKoFyS80gz1fg==[/tex]不随[tex=0.714x1.286]LA74ioWWkXdGbHCtFk/Sog==[/tex]的增加而增加

已知某构件承受轴向拉力设计值[tex=4.429x1.0]ecpHnw20jCaHqR/EZ8dKOg==[/tex]弯矩[tex=5.857x1.0]h688l0du6KT7acVaEv40zQ==[/tex],混凝土强度等级为C30,采用HRB400级钢筋。柱截面尺寸为[tex=18.143x1.357]6VZ4zE9OzDP7sLxnXQk33DGgkgF1l8n0IybE7j30WjpEgbfe/9p/ckL0fd8xgL5oJD6Qu8wYShXmUKsZdXGmo28NLkZLjmtaPEebDRCXbFGsR8oB2Rj/AKcZBdvtycpm[/tex]求所需纵筋面积。

下列变量组（）是一个闭回路。 A: {x，x，x，x，x，x} B: {x，x，x，x，x} C: {x，x，x，x，x，x} D: {x，x，x，x，x，x}

以下谓词蕴含式正确的是（）: (∀x) (A(x)→B(x))=>( ∀x)A(x)→(∀x)B(x)|(∀x) (A(x)↔B(x))=>( ∀x)A(x)↔(∀x)B(x)|(∀x)A(x)∨(∀x)B(x)=>( ∀x) (A(x)∨B(x))|(∃x) (A(x)∧B(x))=>(∃x)A(x)∧(∃x)B(x)

以下谓词蕴含式正确的是（）: (?x) (A(x)→B(x))=>( ?x)A(x)→(?x)B(x)|(?x) (A(x)?B(x))=>( ?x)A(x)?(?x)B(x)|(?x)A(x)∨(?x)B(x)=>( ?x) (A(x)∨B(x))|(?x) (A(x)∧B(x))=>(?x)A(x)∧(?x)B(x)

下列式中错误的是： A: (∀x)(A(x)Úp(x)) Û (∀x)A(x)Ú (∀x)p(x) B: ($x)A(x) Ù p Û ($x)(A(x) Ù p ) C: (∀x)(A(x)ÚB(x)) Þ (∀x)A(x)Ú( ∀x)B(x) D: ($x)(A(x)ÙB(x)) Þ ($x)A(x)Ù( $x)B(x)

判断下列推证是否正确。 (∀x)(A(x)→B(x))⇔(∀x)(￢A(x)∨B(x)) ⇔(∀x)￢( A(x)∧￢B(x) ) ⇔￢(∃x) ( A(x)∧￢B(x) ) ⇔￢( (∃x)A(x)∧(∃x)￢B(x) ) ⇔￢(∃x)A(x)∨￢(∃x)￢B(x) ⇔￢(∃x)A(x)∨(∀x)B(x) ⇔(∃x)A(x)→(∀x)B(x)