20 câu hỏi
Giải hệ phương trình tuyến tính \(\left\{ {\begin{array}{*{20}{c}}{2{x_1} + {x_2} + 2{x_3} + 3{x_4} = 2}\\{6{x_1} + 2{x_2} + 4{x_3} + 5{x_4} = 3}\\{6{x_1} + 4{x_2} + 8{x_3} + 13{x_4} = 9}\\{4{x_1} + {x_2} + {x_3} + 2{x_4} = 1}\end{array}} \right.\)
\[{{\rm{x}}_1} = - 1 - 8{{\rm{x}}_4},{{\rm{x}}_3} = 0,{{\rm{x}}_2} = 1 + 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = - 1 - 8{{\rm{x}}_4},{{\rm{x}}_3} = 1,{{\rm{x}}_2} = 1 + 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = - 1 - 8{{\rm{x}}_1},{{\rm{x}}_3} = 1,{{\rm{x}}_2} = 1 + 2{{\rm{x}}_1}\]
\[{{\rm{x}}_1} = - 1 - 8{{\rm{x}}_1} + 2{{\rm{x}}_2},{{\rm{x}}_4} = 1 + 2{{\rm{x}}_1} - 5{{\rm{x}}_2}\]
Giải hệ phương trình tuyến tính\(\left\{ {\begin{array}{*{20}{c}}{2{x_1} - {x_2} + 3{x_3} + 4{x_4} = 5}\\{4{x_1} - 2{x_2} + 5{x_3} + 6{x_4} = 7}\\{6{x_1} - 3{x_2} + 7{x_3} + 8{x_4} = 9}\\{3{x_1} - 4{x_2} + 9{x_3} + 10{x_4} = 11}\end{array}} \right.\)
\[{{\rm{x}}_1} = 1,{{\rm{x}}_2} = 3 - 2{{\rm{x}}_4},{{\rm{x}}_3} = 4 - 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 0,{{\rm{x}}_2} = 4 - 2{{\rm{x}}_4},{{\rm{x}}_3} = 3 - 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 1,{{\rm{x}}_2} = 3 - 2{{\rm{x}}_3},{{\rm{x}}_3} = 4 + 2{{\rm{x}}_3}\]
\[{{\rm{x}}_1} = 3 + 5{{\rm{x}}_4},{{\rm{x}}_2} = 4,{{\rm{x}}_3} = 3 - {{\rm{x}}_4}\]
Cho hệ phương trình tuyến tính\(\left\{ {\begin{array}{*{20}{c}}{{x_1} + 4{x_2} - 5{x_3} + 9{x_4} = 1}\\{3{x_1} + 2{x_2} + 5{x_3} + 2{x_4} = 3}\\{2{x_1} + 2{x_2} + 2{x_3} + 3{x_4} = 2}\\{2{x_1} + 3{x_2} + 4{x_3} + 2{x_4} = 5}\end{array}} \right.\)
\[{{\rm{x}}_1} = 2,{{\rm{x}}_2} = 3,{{\rm{x}}_3} = - 1,{{\rm{x}}_4} = - 2\]là một nghiệm của hệ
\[{{\rm{x}}_1} = \frac{1}{7},{{\rm{x}}_2} = \frac{{15}}{7},{{\rm{x}}_3} = 0,{{\rm{x}}_4} = \frac{{ - 6}}{7}\]là một nghiệm của hệ
\[{{\rm{x}}_1} = - 11,{{\rm{x}}_2} = - 3,{{\rm{x}}_3} = 6,{{\rm{x}}_4} = 6\]là một nghiệm của hệ
Các trường hợp trên đều là nghiệm của hệ
Giải hệ phương trình tuyến tính\(\left\{ {\begin{array}{*{20}{c}}{2{x_1} + 7{x_2} + 3{x_3} + {x_4} = 5}\\{{x_1} + 3{x_2} + 5{x_3} - 2{x_4} = 3}\\{{x_1} + 5{x_2} - 9{x_3} + 8{x_4} = 1}\\{5{x_1} + 18{x_2} + 4{x_3} + 5{x_4} = 12}\end{array}} \right.\)
\[{{\rm{x}}_1} = 2 + {{\rm{x}}_3} + 7{{\rm{x}}_4},{\rm{x}}2 = - 1 + 5{{\rm{x}}_3} - {{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 6 - 26{{\rm{x}}_3} + 17{{\rm{x}}_4},{\rm{x}}2 = - 1 + 7{{\rm{x}}_3} - 5{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 2 + 6{{\rm{x}}_3} - 7{{\rm{x}}_4},{\rm{x}}2 = - 1 + 4{{\rm{x}}_3} - 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 4 + 11{{\rm{x}}_4},{\rm{x}}2 = - 1 - 6{{\rm{x}}_4},{{\rm{x}}_3} = 2\]
Giải hệ phương trình tuyến tính\(\left\{ {\begin{array}{*{20}{c}}{2{x_1} - {x_2} + {x_3} - {x_4} = 3}\\{4{x_1} - 2{x_2} - 2{x_3} + 3{x_4} = 2}\\{2{x_1} - {x_2} + 5{x_3} - 6{x_4} = 1}\\{2{x_1} - {x_2} - 3{x_3} + 4{x_4} = 5}\end{array}} \right.\)
\[{{\rm{x}}_1} = 2 + 2{{\rm{x}}_4},{{\rm{x}}_2} = 3 - 2{{\rm{x}}_4},{{\rm{x}}_3} = 4 - 2{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = 0,{{\rm{x}}_2} = 1 + 7{{\rm{x}}_4},{{\rm{x}}_3} = - 2 - 5{{\rm{x}}_4}\]
\[{{\rm{x}}_1} = - 4,{{\rm{x}}_2} = - 6 + 3{{\rm{x}}_3},{{\rm{x}}_4} = 7 - 9{{\rm{x}}_3}\]
Hệ vô nghiệm
\[[\forall ({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{),(}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{y}}_2}) \in {R^2}\], biểu thức nào sau đây của η xác định một dạng song tuyến tính của không gian véc tơ R2:
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 2}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}}{\rm{ + 13}}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}} - {\rm{5}}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 3}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}{\rm{ + 2}}{{\rm{x}}_{\rm{2}}}{{\rm{y}}_{\rm{1}}}{\rm{ + 3}}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 2}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}} - {\rm{4}}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}{\rm{ + 4}}{{\rm{x}}_{\rm{2}}}{{\rm{y}}_{\rm{1}}} - {\rm{2}}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = }}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{{\rm{y}}_{\rm{1}}} - {\rm{3}}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[\forall ({{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}) \in \]dạng song tuyến tính η nào sau đây của không gian véc tơ R2 là một tích vô hướng:
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 2}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}}{\rm{ + 8}}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}} - {{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 3}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}}{\rm{ + 8}}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}{\rm{ + 2}}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}{\rm{ + }}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
\[{\rm{\eta ((}}{{\rm{x}}_{\rm{1}}}{\rm{, }}{{\rm{y}}_{\rm{1}}}{\rm{), (}}{{\rm{x}}_{\rm{2}}}{\rm{, }}{{\rm{y}}_{\rm{2}}}{\rm{)) = 2}}{{\rm{x}}_{\rm{1}}}{{\rm{x}}_{\rm{2}}}{\rm{ + }}{{\rm{x}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}} - {\rm{3}}{{\rm{x}}_{\rm{2}}}{{\rm{y}}_{\rm{1}}}{\rm{ + }}{{\rm{y}}_{\rm{1}}}{{\rm{y}}_{\rm{2}}}\]
Tính \[\smallint \cos {\rm{xcos2xdx}}\]
\[ - \frac{1}{6}\cos 3{\rm{x}} + \frac{1}{2}\cos {\rm{x}} + {\rm{C}}\]
\[\frac{2}{3}{\cos ^{\rm{3}}}{\rm{x + cosx + C}}\]
\[ - \frac{2}{3}{\sin ^3}{\rm{x + sinx + C}}\]
Đáp án A và C đều đúng
Tính \[\smallint {(1 + 2{\rm{x)}}^{{\rm{2013}}}}{\rm{dx}}\]
\[\frac{1}{{4028}}{(1 + 2{\rm{x)}}^{{\rm{2013}}}}{\rm{ + C}}\]
\[\frac{1}{2}{(1 + 2{\rm{x)}}^{{\rm{2013}}}}{\rm{ + C}}\]
\[\frac{1}{{4024}}{(1 + 2{\rm{x)}}^{{\rm{2013}}}}{\rm{ + C}}\]
\[\frac{1}{{2013}}{(1 + 2{\rm{x)}}^{{\rm{2013}}}}{\rm{ + C}}\]
Tính \[\smallint {\mathop{\rm si}\nolimits} {\rm{n}}\left( {\frac{{\rm{\pi }}}{{\rm{3}}} - \frac{{\rm{x}}}{{\rm{4}}}} \right){\rm{dx}}\]
\[\frac{1}{2}\cos \left( {\frac{{\rm{\pi }}}{{\rm{3}}} - \frac{{\rm{x}}}{{\rm{2}}}} \right){\rm{ + C}}\]
\[4\cos \left( {\frac{{\rm{\pi }}}{{\rm{3}}} - \frac{{\rm{x}}}{{\rm{4}}}} \right){\rm{ + C}}\]
\[2\sin \left( {\frac{{\rm{\pi }}}{{\rm{3}}} - \frac{{\rm{x}}}{{\rm{2}}}} \right){\rm{ + C}}\]
\[\frac{1}{2}\sin \left( {\frac{{\rm{\pi }}}{{\rm{3}}} - \frac{{\rm{x}}}{{\rm{2}}}} \right){\rm{ + C}}\]
Tính\[\smallint \cot 5{\rm{xdx}}\]
\[ - \frac{1}{3}\ln |\cos {\rm{3x| + C}}\]
\[\frac{1}{3}\ln |\cos 5{\rm{x| + C}}\]
\[ - \frac{1}{3}\ln |\sin 3{\rm{x| + C}}\]
\[\frac{1}{5}\ln |\sin 5{\rm{x| + C}}\]
Tính tích phân \[{\rm{I}} = \smallint \frac{{{\rm{3dx}}}}{{{{\rm{x}}^2} - 7{\rm{x}} + 10}}\]
\[\ln |{\rm{x}} - 2| - \ln |{\rm{x}} - 4| + {\rm{C}}\]
\[\ln |{\rm{x}} - 5| - \ln |{\rm{x}} - 2| + {\rm{C}}\]
\[\frac{{\ln |{\rm{x}} - 5|}}{{\ln |{\rm{x}} - 2|}} + {\rm{C}}\]
\[\ln |({\rm{x}} - 4)(2 - 2)| + {\rm{C}}\]
Tính tích phân \[{\rm{I}} = \smallint \frac{{7{{(\ln {\rm{x}} - 1)}^6}}}{{\rm{x}}}{\rm{dx}}\]
\[\frac{{{{\ln }^3}{\rm{x}} - 2\ln {\rm{x}} + 1}}{{{{\rm{x}}^2}}} + {\rm{C}}\]
\[{(\ln {\rm{x}} - 1)^7} + {\rm{C}}\]
\[{\ln ^3}{\rm{x}} - 2\ln {\rm{x}} + 1 + {\rm{C}}]\]
Tính\[\smallint \frac{{{\rm{dx}}}}{{\sqrt[{\rm{3}}]{{{{{\rm{(5x + 3)}}}^{\rm{2}}}}}}}\]
\[\frac{3}{5}\sqrt[{\rm{3}}]{{{\rm{5x + 3}}}}{\rm{ + C}}\]
\[ - \frac{3}{2}\sqrt[{\rm{3}}]{{{\rm{2x + 3}}}}{\rm{ + C}}\]
\[\sqrt[{\rm{3}}]{{{\rm{2x + 3}}}}{\rm{ + C}}\]
\[\frac{1}{2}\sqrt[{\rm{3}}]{{{\rm{5x + 3}}}}{\rm{ + C}}\]
Tính\[\smallint \frac{{{\rm{dx}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{(}} - {\rm{3x + 1)}}}}\]
\[\frac{1}{3}\cot ( - 3{\rm{x}} + 1) + {\rm{C}}\]
\[\frac{1}{2}\tan ( - 2{\rm{x}} + 1) + {\rm{C}}\]
\[ - \frac{1}{3}\cot ( - 3{\rm{x}} + 1) + {\rm{C}}\]
\[ - \frac{1}{2}\tan ( - 2{\rm{x}} + 1) + {\rm{C}}\]
Tính \[\smallint \frac{{{\rm{2}}{{\rm{e}}^{\rm{x}}}{\rm{dx}}}}{{{{\rm{e}}^{{\rm{2x}}}} - {\rm{2}}{{\rm{e}}^{\rm{x}}}{\rm{ + 1}}}}\]
\[\frac{2}{{{{\rm{e}}^{\rm{x}}} - 1}} + {\rm{C}}\]
\[ - \frac{2}{{{{\rm{e}}^{\rm{x}}} - 1}} + {\rm{C}}\]
\[ - \frac{{{{({{\rm{e}}^{\rm{x}}} - 1)}^3}}}{3} + {\rm{C}}\]
\[\frac{{{{({{\rm{e}}^{\rm{x}}} - 1)}^3}}}{3} + {\rm{C}}\]
Tính tích phân xác định\[{\rm{I}} = \mathop \smallint \limits_1^{\rm{e}} \frac{{{\rm{dx}}}}{{{\rm{(2x(1 + l}}{{\rm{n}}^{\rm{2}}}{\rm{x)}}}}\]
\[\frac{{\rm{\pi }}}{{\rm{8}}}\]
\( - \frac{{\rm{\pi }}}{{\rm{8}}}\)
\(\frac{\pi }{2}\)
1
Tính tích phân xác định\[{\rm{I}} = \mathop \smallint \limits_1^{\rm{e}} 8{\rm{xlnxdx}}\]
2
\[{{\rm{e}}^2} - 1\]
\[2{{\rm{e}}^1} + 2\]
e
Tính tích phân xác định\[{\rm{I}} = \mathop \smallint \limits_{ - 2}^0 \frac{{{\rm{3xdx}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + 2x + 2}}}}\]
\[\frac{{{\rm{3\pi }}}}{{\rm{2}}}\]
\(\frac{\pi }{4}\)
1
0
Tính tích phân xác định\[{\rm{I}} = \mathop \smallint \limits_{\frac{{\rm{\pi }}}{{\rm{6}}}}^{\frac{{\rm{\pi }}}{{\rm{3}}}} 4\cot {\rm{xdx}}\]
2ln2
2ln3
-1
1