A. \[\overrightarrow {{u_1}} = \left( {{\rm{1}};0} \right)\];
B. \[\overrightarrow {{u_2}} = \left( {0; - {\rm{1}}} \right);\]
C. \[\overrightarrow {{u_3}} = \left( { - {\rm{1}};1} \right);\]
D. \[\overrightarrow {{u_4}} = \left( {{\rm{1}};{\rm{1}}} \right).\]
A. \[\overrightarrow {{u_1}} = \left( {1; - 1} \right);\]
B. \[\overrightarrow {{u_2}} = \left( {0;1} \right);\]
C. \[\overrightarrow {{u_3}} = \left( {1;0} \right);\]
D. \[\overrightarrow {{u_4}} = \left( {1;1} \right).\]
A. \[\overrightarrow {{u_1}} = \left( { - 1;2} \right);\]
B. \[\overrightarrow {{u_2}} = \left( {2;1} \right);\]
C. \[\overrightarrow {{u_3}} = \left( { - 2;6} \right);\]
D. \[\overrightarrow {{u_4}} = \left( {1;1} \right).\]
A. \[\overrightarrow {{u_1}} = \left( {0;a + b} \right);\]
B. \[\overrightarrow {{u_2}} = \left( {a;b} \right);\]
C. \[\overrightarrow {{u_3}} = \left( {a; - b} \right);\]
D.\[\overrightarrow {{u_4}} = \left( { - a;b} \right).\]
A. \[\overrightarrow {{n_1}} = \left( {a; - b} \right);\]
B. \[\overrightarrow {{n_2}} = \left( {a;b} \right);\]
C. \[\overrightarrow {{n_3}} = \left( {b;a} \right);\]
D. \[\overrightarrow {{n_4}} = \left( { - b;a} \right).\]
A. \(1\);
B. \(2\);
C. \(4\);
D. Vô số.
A. \[d:\left\{ \begin{array}{l}x = 3 + t\\y = 5 - 2t\end{array} \right.\];
B. \[d:\left\{ \begin{array}{l}x = 1 + 3t\\y = - 2 + 5t\end{array} \right.\];
C. \[d:\left\{ \begin{array}{l}x = 1 + 5t\\y = - 2 - 3t\end{array} \right.\];
D. \[d:\left\{ \begin{array}{l}x = 3 + 2t\\y = 5 + t\end{array} \right.\].
A. \[d:\left\{ \begin{array}{l}x = - 1\\y = 2\end{array} \right.\];
B. \[d:\left\{ \begin{array}{l}x = 2t\\y = t\end{array} \right.\];
C. \[d:\left\{ \begin{array}{l}x = t\\y = - 2t\end{array} \right.\];
D. \[d:\left\{ \begin{array}{l}x = - 2t\\y = t\end{array} \right.\].
A. \[{d_1}:\left\{ \begin{array}{l}x = 3 + 2t\\y = t\end{array} \right.\];
B. \[{d_2}:\left\{ \begin{array}{l}x = - t\\y = - 2 + 3t\end{array} \right.\];
C. \[{d_3}:\left\{ \begin{array}{l}x = 3 + t\\y = - 2t\end{array} \right.\];
D. \[{d_4}:\left\{ \begin{array}{l}x = 3t\\y = - 2\end{array} \right.\].
A.\[\overrightarrow {{u_1}} = \left( {6;0} \right)\];
B.\[\overrightarrow {{u_2}} = \left( { - 6;0} \right)\];
C.\[\overrightarrow {{u_3}} = \left( {2;6} \right)\];
D.\[\overrightarrow {{u_4}} = \left( {0;1} \right)\].
A. y + 2 = 0;
B. x + 1 = 0;
C. x - 1 = 0;
D. y - 2 = 0.
A. d :\[\left\{ \begin{array}{l}x = 10 + t\\y = 6\end{array} \right.\];
B. \[d:\left\{ \begin{array}{l}x = 2 + t\\y = - 10\end{array} \right.\];
C. \[d:\left\{ \begin{array}{l}x = 6\\y = - 10 - t\end{array} \right.\];
D. \[d:\left\{ \begin{array}{l}x = 6\\y = - 10 + t\end{array} \right.\].
A. -x + 3y + 6 = 0 ;
B. 3x - y + 10 = 0 ;
C. 3x - y + 6 = 0 ;
D. 3x + y - 8 = 0.
A. 2x - 3y + 4 = 0 ;
B. 3x - 2y + 6 = 0 ;
C. 3x - 2y - 6 = 0 ;
D. 2x - 3y - 4 = 0.
A. x + y - 1 = 0 ;
B. 2x - 7y + 9 = 0 ;
C. x + 2 = 0 ;
D. x - 2 = 0.
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