Đề online: Luyện tập Phương trình tham số và phươn...

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 \(\overrightarrow u  = \left( {6;\,0} \right)\)                

B \(\overrightarrow u  = \left( { - 6;\,0} \right)\)             

C \(\overrightarrow u  = \left( {2;\,6} \right)\)

D \(\overrightarrow u  = \left( {0;\,1} \right)\)

A \(\overrightarrow {{n_1}}  = \left( {2;3} \right)\)      

B \(\overrightarrow {{n_2}}  = \left( { - 2;3} \right)\)                           

C \(\overrightarrow {{n_3}}  = \left( {2;\, - 1} \right)\)                         

D \(\overrightarrow {{n_4}}  = \left( { - 2;\,\,1} \right)\)   

A \(\overrightarrow {{u_1}}  = \left( {1;\,\,2} \right)\)

B \(\overrightarrow {{u_2}}  = \left( {2;\,4} \right)\)    

C \(\overrightarrow {{u_3}}  = \left( {\frac{1}{2};\,1} \right)\)

D \(\overrightarrow {{u_4}}  = \left( {2;\,\, - 4} \right)\)

A \(d:\,\,\frac{{x - 3}}{2} = \frac{y}{1}\)

B \(d:\,\,\frac{x}{1} = \frac{{y + 2}}{3}\)                    

C \(d:\,\,\frac{x}{3} = \frac{y}{{ - 2}}\)          

D \(d:\,\,\frac{x}{3} = \frac{{y + 2}}{1}\)

A \(\frac{{x + 2}}{1} = \frac{{y - 1}}{0}\)

B \(\frac{{x - 2}}{2} = \frac{{y + 1}}{{ - 1}}\)                      

C \(\frac{{x - 2}}{{ - 1}} = \frac{{y + 1}}{2}\)

D \(\frac{{x + 2}}{0} = \frac{{y - 1}}{1}\)

A \(\left\{ \begin{array}{l}x = 4 + 5t\\y = 5 + 3t\end{array} \right..\)

B \(\left\{ \begin{array}{l}x = 4 - 5t\\y = 5 + 3t\end{array} \right..\)   

C \(\left\{ \begin{array}{l}x =  - 4 + 5t\\y =  - 5 + 3t\end{array} \right..\)                    

D \(\left\{ \begin{array}{l}x = 6 + 5t\\y = 1 + 3t\end{array} \right..\)

A \(\left\{ \begin{array}{l}x = 2\\y =  - 3 - t\end{array} \right.\)           

B \(\left\{ \begin{array}{l}x = 2 + t\\y =  - 3 + t\end{array} \right.\)    

C \(\left\{ \begin{array}{l}x = 2\\y =  - 3 + t\end{array} \right.\)          

D \(\left\{ \begin{array}{l}x = 2 + t\\y =  - 3\end{array} \right.\)

A \(\frac{{x - 1}}{3} = \frac{{y + 3}}{{ - 6}}\)          

B \(\frac{{x + 1}}{1} = \frac{{y - 3}}{{ - 2}}\)                      

C \(\frac{{x - 1}}{1} = \frac{{y + 3}}{{ - 2}}\)          

D \(\frac{{x + 1}}{{ - 1}} = \frac{{y - 3}}{{ - 2}}\)

A \(y = \sqrt 3 x + 1\)                   

B \(y = \sqrt 3 x - 5\)                   

C \(y = \sqrt 3 x + 5\)      

D \(y = \sqrt 3 x - 1\)

A \(\left\{ \begin{array}{l}x = 5t\\y = 3 + t\end{array} \right.\)            

B \(\left\{ \begin{array}{l}x = 5\\y = 1 + 3t\end{array} \right.\)           

C \(\left\{ \begin{array}{l}x = t\\y = 3 - 5t\end{array} \right.\)             

D \(\left\{ \begin{array}{l}x = 3 + 5t\\y = t\end{array} \right.\)

A \(\frac{{x - 1}}{{\sqrt 2 }} = \frac{{y + \sqrt 2 }}{{ - 1}}\)

B \(\frac{{x + 1}}{{2\sqrt 2 }} = \frac{{y - \sqrt 2 }}{{ - 1}}\)

C \(\frac{{x + 1}}{{\sqrt 2 }} = \frac{{y - \sqrt 2 }}{{ - 1}}\)                        

D \(\frac{{x - 1}}{{2\sqrt 2 }} = \frac{{y + \sqrt 2 }}{{ - 1}}\)

A \(\frac{{x - 3}}{4} = \frac{{y - 2}}{{ - 2}}\)

B \(\frac{{x - 3}}{2} = \frac{{y - 2}}{1}\)                  

C \(\frac{{x - 3}}{{ - 2}} = \frac{{y - 2}}{1}\)

D \(\frac{{x + 1}}{2} = \frac{{y + 2}}{1}\)

A \(2\)      

B \( - 2\)

C \(1\)

D \(-1\)

A \(\left\{ \begin{array}{l}x =  - 3 + t\\y = 5 - t\end{array} \right.\)

B \(\left\{ \begin{array}{l}x =  - 3 + t\\y = 5 + t\end{array} \right.\)    

C \(\left\{ \begin{array}{l}x = 3 + t\\y =  - 5 + t\end{array} \right.\)

D \(\left\{ \begin{array}{l}x = 5 - t\\y =  - 3 + t\end{array} \right.\)

A \(\left\{ \begin{array}{l}x = 4 - t\\y =  - 7 - t\end{array} \right.\)              

B \(\left\{ \begin{array}{l}x = 4\\y =  - 7 + t\end{array} \right.\)          

C \(\left\{ \begin{array}{l}x = 4 + t\\y =  - 7 + t\end{array} \right.\)    

D \(\left\{ \begin{array}{l}x = 4 + t\\y =  - 7\end{array} \right.\)

A \(\frac{{x + 1}}{{ - 2}} = \frac{{y - 3}}{{ - 1}}\)   

B \(\frac{{x + 1}}{2} = \frac{{y - 3}}{{ - 1}}\)          

C \(\frac{{x - 1}}{2} = \frac{{y + 3}}{{ - 1}}.\)

D \(\frac{{x + 1}}{2} = \frac{{y - 3}}{1}.\)

A \(k = 1\)                       

B \(k = 2\)           

C \(k =  - 1\)

D \(k =  - 2\)

A \(\left\{ \begin{array}{l}x = 7\\y = 3 + 5t\end{array} \right.\)

B \(\left\{ \begin{array}{l}x = 3 - 5t\\y =  - 7\end{array} \right.\)         

C \(\left\{ \begin{array}{l}x = 7 + t\\y = 3\end{array} \right.\)                         

D \(\left\{ \begin{array}{l}x = 2\\y = 3 - t\end{array} \right.\)

A \(\left\{ \begin{array}{l}x =  - 2 + 3t\\y =  - 2 - 2t\end{array} \right.\)

B \(\left\{ \begin{array}{l}x =  - 2 - 4t\\y = 1 - 3t\end{array} \right.\)

C \(\left\{ \begin{array}{l}x =  - 2 - 3t\\y = 1 - 4t\end{array} \right.\)

D \(\left\{ \begin{array}{l}x =  - 2 - 3t\\y = 1 + 4t\end{array} \right.\)