已知矩阵方程[tex=12.714x2.786]X4MGaXJtau0+12XWdlTCIA9wFciqbGIfGw4Ld8J4V7ePT85XuNAjYv2sdu1648UkfX+F0BQnM3bAymN5xXBja6MxRDWxH12oIi97C3bfY+NxB1hq9AW6wtAGoehfwIL6N6IPGHKVMpWHX/G7+v8zcL6GXBpoYmQ+bne3E0WY9wg=[/tex]，求矩阵[tex=0.857x1.0]KGogyvwDAIJf/iL0H/9wjg==[/tex].

已知矩阵方程[tex=12.714x2.786]X4MGaXJtau0+12XWdlTCIA9wFciqbGIfGw4Ld8J4V7ePT85XuNAjYv2sdu1648UkfX+F0BQnM3bAymN5xXBja6MxRDWxH12oIi97C3bfY+NxB1hq9AW6wtAGoehfwIL6N6IPGHKVMpWHX/G7+v8zcL6GXBpoYmQ+bne3E0WY9wg=[/tex]，求矩阵[tex=0.857x1.0]KGogyvwDAIJf/iL0H/9wjg==[/tex].

 求下列两个信道的容量，并加以比较。[tex=12.714x2.786]BJ01inipONhLpj7h0EHN8iUJwpdn3WhKSrpQLMQ8Dedpy6NkcXgUE76DaH9/MZuM3TjOYX6y/vvReOjKx1KE/U96byxeeaZQhhMnKXJKDwxpqUcL1/afvK1pns1nX3Yw[/tex] 

已知矩阵方程[tex=12.714x2.786]X4MGaXJtau0+12XWdlTCIA9wFciqbGIfGw4Ld8J4V7ePT85XuNAjYv2sdu1648UkfX+F0BQnM3bAymN5xXBja6MxRDWxH12oIi97C3bfY+NxB1hq9AW6wtAGoehfwIL6N6IPGHKVMpWHX/G7+v8zcL6GXBpoYmQ+bne3E0WY9wg=[/tex]，则[tex=1.643x1.0]priNy5ZwzZYuFo1ohjDY5Q==[/tex][input=type:blank,size:4][/input]。

 指出下列集合中哪些是 [tex=3.0x1.357]BTuMgdu50zk9vvD7VuEqDc/wzH/S0fh+6wscfZT1CSo=[/tex]的子环?(1) [tex=13.071x2.786]cc3WTj0JH/gK5NTGIlwtOkLAGtZRyyhWRnOlEgS2ZEJjlz4LcsDGHMBEzMBXr+Yc8ysYINaJpZgzBx2x2+5yBK6sizDU8wobe4ZOIE9QejaehVT3/YUABpZdJqiVMGytAyWxyfPPFciKrsEpO0ReJg==[/tex](2)[tex=12.714x2.786]cc3WTj0JH/gK5NTGIlwtOkLAGtZRyyhWRnOlEgS2ZELTWCXQYbcHfOtXa8OQ1hnd7bi5XfEyENlNqz81Q9nHow0SaOBUe9ZesZIzFTec1FNMp0Yd7WjIdMVfFUdkLsbWHSula8qf2X/g7ovxTdJfgA==[/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)