[tex=37.143x1.286]P5foZyov7jGafq+n3G+hH5vcSAVdTaSdoTfHRJ4FiyKKNzyFNgOhJjc6bV47n4t91CpLJyMXHoBp2+v/2+wdLyHAhixxQjNTz0sQ7kEcSprNqynC+pWOjNbT/MSzRMEKYRX4bp4qbaxlhnUOuZ7guWXl/QzJjZHID9T7ylrjKro=[/tex]

[tex=37.143x1.286]qqaLp0IAbzxZZ6DjxRmth3w5F64OJGSNArOMPgimE5e+tqFYCTGX4vVz5o5RO85VrMh4PWjM2gLH9jQRpDmT+dbsKeahugN6eAJLWGXeyyqcSE/km9KpPprJx1TWWMnvqM24Qcvj2GqMFOIdEqUcl+o5VUuELTR/c1M3/gqfN9w=[/tex][tex=8.643x1.286]mvjS7PQtVtBXG2I49pXNVtgJeFpCc7MxNKErs8ukeBomf3MOuiHzOqkEfdOm98bn[/tex]

下列选项为某变速器一、二、三、四挡的传动比，则三挡的传动比为（）。 A: 0.969 B: 1.286 C: 1.944 D: 3.455

[tex=32.786x1.286]ZCdaOYI7cJqUAJoluXWiMPebJkycPvgcru0+q+Qrx0mCjxCWgLd+iXHnG6EYlFNSAxxNZ6viNxAizZwhc6RMEXdl0TuhzXPCFiWotjycH7+xoikLvHnTBgqSeGQoX27yZPyVf5NjwVrCFK09Ew17xHK4B41Bb23Jh4GURq+4AAI=[/tex][tex=40.357x1.286]DNO1tPWBDXKMtaDRuVW8rbfL5k0ig4LNrOQOZjYnm3T7S+km3ZA3PybY1+E7hu7IrjC/m+pHSRu9pX8JySaq6YMkhaxL2IMRgPzPUlaqiHdGw4tc/CYCG7jgXwgC37dpQz1f/PWGkMeWul5BU7WNzsuvSUXt7P/IogVF14wKBbh7yoY9n82NMPhCGEi7Ogxj[/tex][tex=37.143x1.286]dwtKSqYTFeO8lk+zkipT9z6xgQUlYwkCyv+1xo56Ez7E9Rz+KeBem0W5a+DczInKkUL6uGfax32vV2Z8OlXZJj2R9nJHkGPFR3S6UxqN/GCgAEKOjIhMiycXhFikiuIlOmjUFQY4kLweDAxSstbgRp1mJce0y6EjmUo4+QhYAww1UmQE9PhdAAgN5/1fQRz3[/tex][img=922x129]17a36bcd478a97e.png[/img]

下面是某立方晶系物质的几个晶面，试将它们的面间距从大到小按次序重新排列: [tex=37.143x1.286]qfMXmuVjgmVhQ//kulYmiEQ/5nGYe1fNAYOYoqs1y5vKxhmLpvd7NtqxUG4xnIORcx91CcuZRXyoMHsNSH8xM5s7NoCk2Y51Zu0Bz/NlS2Ym3pzu1MU+KU3LQ6YHth+DLRT8JdGgmxJcr0hHIlbsdQ==[/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)