设 [tex=19.071x1.357]9iT23czbHhQCES+iSNK3EWjNyDqC+/sEar1vc88p9dr2wcSf7sMmOepiRO4q42/qgGhedcwB9YuHA56cMEAIyw==[/tex], 求[tex=0.5x0.786]gdMkE6SnyZedYLxpUxdkaQ==[/tex].

已知[tex=27.143x3.643]sSXBpxJWudVpH1R35o4LnNWgMgW7arJ/wkhZv1J4gdigMZA8SWC/I/b5eA+eok2mw7O9IAlqy3uHSOFiNQMhJz5BzDhVjKPGG69VdZP1SsAZn8g/qwFWHaRylraffno5044OAW9b9wezr9otXovo+tr29dv8NArCJ1XfT9EZmjZscjWpmm+xSehEtdUjijtWuIL5/IN45mSqo251lVNFKsjrdhiAYGXAopZJzTDUDtvBQcIqilCSuDjluA57GiJAoU49oW+9KMzmr/3R3znRmTz7gpACvcELjG3MnsPvX6E=[/tex]求[tex=19.071x1.357]SwHq7TpS1Oc08cmPOem8i1qfCFLf+nuJu+6L/AucAZR7cntgXvwv2OVbmCTWn2qk0TFHTUDEdeVIzF7qIzJc16mnb+ee7Yf7x9Fwl8BD9GIqig/V28ZgHotO0MKUiCrQPF/qm3/4hAUc/m5pJvkMy8nN8JD63oTxmKexWAlKY1iI6ZByTW3xZR1d1yL++tjna45Rd/O8PlxbGwNlJItT7SKwo1FZglb15rc4GCDXXQ9jgDUHW4xyysZYo/F6GJzWA6uLuny+gIJrczEo4yk6qJPe3fcdsHI6UfyJQBkR164=[/tex]

求下列差分方程所描述的离散系统的零输入响应,零状态响应和全响应。[tex=19.071x1.357]zVtlD8wh3ZsJinYEiFPSrYwjUZ7xZeznNDrVSuJmROhzc3BkHp3P261cvCwaMSa7wpaunPBjOAt3+NTvesi3mw==[/tex]

设[tex=2.5x1.357]u6Xfo9ZsA1MrVpfnGabm9QipQeeksSHm1ZlBtCHi0pFyyZ+Hjqp4mKEuP9x1cxW2[/tex]是格，证明：对任意的a、b、c∈L，则[tex=19.071x1.357]4NHwUC9+dZ5xihpXZbvxSoSR/U3gshq5QuXeWtrmIHH9t2aSCfVOPHjs7VhBHFnaZnrBWvT9u5N/1D+K0sDLLVChi/OWDk1X9G/rpIsbA6hLIWLp+ZyaioRLQwGaYDHJ[/tex]。

求下列差分方程所描述的[tex=1.714x1.0]ptAt65AfwA4koryK6UNdFg==[/tex]离散系统的零输入响应、零状态响应和全响应。[br][/br][tex=19.071x1.357]m42gBaifKux86I+xVF46Xej5ZiUA9t6Vb7HI3Wn+xu6mZRXQXnc4sAVqRKOe+EZG[/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)