求下列函数的傅氏积分[p=align:center][tex=13.286x3.357]yA9VxGgWFghS/CdxQGqmXWUzsxgUaTrmcUE4ie0vHWlQaHc+dpo8ZwTt5J23rytNnSw53x8AuoyPc3UQEWec4kjnQ5O03Ab6Lm5PDrJy3WQ24hgQ/VG8zSjmiFfGrjuzvRItzhIbfkv80VESoHpPnG/DiHg61oibuURywTQpnY8=[/tex]

确定下列函数的定义域，并求常数[tex=0.571x0.786]7G1MINzwputr5mgALyjQfA==[/tex]，[tex=0.429x1.0]JThLUuJ8WswSAPiYZWihWg==[/tex]使函数在定义域内连续：[tex=13.286x3.357]0Oc6OdDyTxw5ASPscCgHyQx+T3TLizBSA8qXRCV22SIJtYjTnEV+b+9Xl+kcN3VVAsT2hlXkDnX3UIwyjEg9LDg/ZiqZAUtByMBIYcM0q5cP1NVHGFmo9gDpR11qG7qx[/tex]

图示压杆体系中，弹簧支座刚度为[img=64x44]180379b3117202b.png[/img]，则该体系的临界荷载为kl的（ ）倍。[img=3924x1660]180379b32884d13.jpg[/img] A: 0.411 B: 1 C: 1.645 D: 3.357

图示压杆体系中，弹簧支座刚度为[img=64x44]1802df48734317e.png[/img]，则该体系的临界荷载为kl的（ ）倍。[img=3924x1660]1802df4888c0309.jpg[/img] A: 0.411 B: 1 C: 1.645 D: 3.357

 证明 : 由方程组[tex=13.286x3.357]GE56u9QCDTqcLxZ66HADyohtASTr8TjP4SCKA7okDAhsxc6tKch7/CuHiKKnO6jjTm8zYtHnEmPlR8RCZGfydTdYrL1FdM9U99aMh7nmOCfRW9if89Z6Q9YemVu1ew9u4Lq4jFVJ0bKDcSzYwdufQnZSGhonnpN/ho1MqS0k6Pofa9XusELdgG7Zm4sa9SKw[/tex][其中 [tex=4.357x1.357]5NTiU76VyU0F8oYQat0EDE2YjgyD7RH/F2DW0RUe+W0=[/tex] 为参变量,[tex=1.929x1.357]OuWeBotLOyA9BpzktnAd3Q==[/tex] 为任意可微函数]定义的函数[tex=4.0x1.357]AcsG/g4yU933TGrJs0qkCQ==[/tex] 满足方程[tex=9.286x2.929]PYPlEgwNKA9y6IkAp7Vv1FwGu45nlgrUetlOcJk2EbFCoobCLamHjJDJi0aRuyPHCnGeIza9xQNfiJHootAQX5hs2UIcbONsY7QsY/LZXPbN3lCMIN+kd7ftRucG7wE9L1Z08bhZp6i0mWyIfsGA7Q==[/tex]

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)