解下面微分方程初值问题.[tex=7.929x4.786]fnpmC2J6JmQBLyo5NmGAz7w1eTyrPAhhi3OxImldbuDKF7PW8cfBvAB2EbWbHAQRfGWkVyKjxcK5JI3sZEqorAyd19mrtY3Mob47Ii1er++pkoohBQS/6f6ekDExwOkF1ZXPI48Bge7mMf/vRsYiuEUDb1IRutddJXgqRR2aBUo=[/tex]

解下面微分方程初值问题.[tex=7.929x4.786]fnpmC2J6JmQBLyo5NmGAz7w1eTyrPAhhi3OxImldbuDKF7PW8cfBvAB2EbWbHAQRfGWkVyKjxcK5JI3sZEqorAyd19mrtY3Mob47Ii1er++pkoohBQS/6f6ekDExwOkF1ZXPI48Bge7mMf/vRsYiuEUDb1IRutddJXgqRR2aBUo=[/tex]

计算下列矩阵的 [tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex] 次幂, 其中 [tex=0.571x1.0]rFc/sfAAuCOtzhevhoREeA==[/tex] 为正整数:[tex=7.929x4.786]SG13E7iu2HdaLVWfWJMdasNcssnOsnpcSXP9pfv8ZVudX8uBxPyIW+BW1iuKqBWPQy19xF0hvC5K+ZJXm49WVAb1VZdwsjQNiE6Ohf5lij4=[/tex]

设4阶行列式[tex=7.929x4.786]ZpOCMnLTWn2edxCRLh4N3S0Ge1qQtW5wwvt/KpGrsT5wB9zt6llRF2hqErzbsfU1Gl/GOCPB+Kr0MtQcxyVYfQIwaQv1dCmoK2qEGUsyQS341Y0jdAklzylsVMOdSyC6[/tex]，则[tex=9.5x1.214]NCIkt9iygT1uZ1GeH2jfNVr1wsdzQR1SmgpsqEkjh/4=[/tex][tex=0.786x0.643]NhuTNiqjImitwKaHFutGOg==[/tex][input=type:blank,size:4][/input]．

设3阶方阵[tex=2.071x1.286]9rhJ1AXzFOKDcBia45pbow==[/tex]满足[tex=8.214x1.286]7ZYjHEE7HzXQJcFtPFXRUgbBcpKqCY9wWbnP7eFBKHw=[/tex]，其中[tex=0.786x1.286]esgwf7UWeyPnizAgk3o6GQ==[/tex]为3阶单位矩阵，若[tex=7.929x4.786]YmZOrAutSxe6/7rdwvN+7x0FYJYvmLYrtEX+ZY463SmT8z4YzKtAYeBgpcaudgY1hBBc5hi0No7es8ISLjNzjultKhvtpN61XqpwJfAR8qo=[/tex]，则[tex=2.357x1.286]O9l4Mx+uOi2e8W+XU8+OqA==[/tex][input=type:blank,size:4][/input]。

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)