设3阶矩阵[tex=7.857x3.5]2vxBFm+jNAMVDxKL9tbIt1eG7q7J/i8cOQMkyMVtRZRvRLzDLF8V6rgQWP1rZx+RhoijlkgLohOgLDsQT5oqEllbkFPxhlrg7bGVq34+eOiyynw/7iemOxXXy1uyWx+9[/tex],P为3阶非零矩阵,且[tex=3.0x1.214]CCJnY66rJw4UYKpIh22t9g==[/tex],则( )
未知类型:{'options': ['t =6时,R(P)= 1', 't=6时,R(P)=2', 't≠6时,R(P)=1', 't≠6时,R(P)=2'], 'type': 102}
未知类型:{'options': ['t =6时,R(P)= 1', 't=6时,R(P)=2', 't≠6时,R(P)=1', 't≠6时,R(P)=2'], 'type': 102}
举一反三
- 利用反证法证明:R∨S,R→¬Q,S→¬Q,P→Q=>¬P请将下面推理论证的过程补充完整。(说明:输入答案时,不要输入多余的空格)证明过程如下:(1)( ) 假设前提 (2)P→Q P(3) Q T(1)(2) I(4)S→¬Q P(5)( ) T(3)(4) I(6)R∨S P(7)R T(5)(6) I(8)R→¬Q P(9)¬Q T(7)(8) I(10)( )矛盾 T(3)(9) I
- 1 Complete the words ► p r e v e n t prevent 1 i n f _ _ _ i o n ____ 2 b _ n d _ g e ____ 3 t h _ r _ _ g h l _ ____ 4 w _ _ n d ____ 5 b l _ _ d ____ 6 s _ m p t _ m ____ 7 t _ _ p o r _ r _ l y ____ 8 d _ v _ l _ p ____
- 已知,P为三阶非零矩阵,且PQ=0,则 A: t=6时,R(P)=1 B: t=6时,R(P)=2 C: t6时,R(P)=1 D: t6时,R(P)=2
- 构造下式的推理证明:有理数都是实数,有的有理数是整数,因此有的实数是整数。证明设Q(x):x为有理数;R(x):x为实数;Z(x):x为整数;前提:∀x(Q(x)→R(x)),∃x(Q(x)⋀Z(x));结论:∃x(R(x)⋀Z(x))。(1)∃x(Q(x)⋀Z(x)) P(2)Q(c)⋀Z(c) ES(1)(3)∀x(Q(x)→R(x)) P(4)Q(c)→R(c) US(3)(5)Q(c) T(2)I(6)R(c) T(2)(4)I(7)Z(c) T(2)I(8)R(c)⋀Z(c) T(6)(7)I(9)∃x(R(x)⋀Z(x)) EG(8)以上推理是有效的。 A: 正确 B: 错误
- 1 Complete the words. ► o u t s t a n d i n g outstanding 1 e_ _ _ v a l e n t ____ 2 _ _ _ d u c e ____ 3 _ _ _ p o r t i n g r o _ _ ____ 4 _ _ _ i v i d u a l ____ 5 a _ _ r d ____ 6 a c _ _ e v e ____ 7 w_ _ n _ r ____ 8 p _ _ z e ____