Câu hỏi:

Ta có: \({{u}_{n}}={{u}_{1}}+\left( n-1 \right)d.\)

\({{u}_{3}}=2+2.\left( -3 \right)=-3\Rightarrow A,B\) sai.

\({{u}_{4}}=2+3.\left( -3 \right)=-7\Rightarrow \) C đúng.

\({{u}_{6}}=2+5\left( -3 \right)=-13\Rightarrow D\) sai.

Chọn C.

Gọi M’(x; y) là ảnh của điểm M qua phép tịnh tiến theo vector \(\overrightarrow{u}\).

\(\begin{array}{l}{T_{\overrightarrow u }}\left( M \right) = M' \Leftrightarrow \overrightarrow {MM'}  = \overrightarrow u \\ \Leftrightarrow \left( {x - 2;y - 3} \right) = \left( { - 2;1} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}x - 2 =  - 2\\y - 3 = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x = 0\\y = 4\end{array} \right. \Rightarrow M'\left( {0;4} \right).\end{array}\)

Chọn B.

Ta có \({{T}_{\overrightarrow{u}}}\left( A \right)=B\Leftrightarrow\overrightarrow{AB}=\overrightarrow{u}\Leftrightarrow \left( 1;-1\right)=\overrightarrow{u}.\)

Chọn B.

x \(\begin{array}{l}\overrightarrow {AB'}  = \frac{1}{2}\overrightarrow {AB}  \Rightarrow {V_{\left( {A;\frac{1}{2}} \right)}}\left( B \right) = B'\\\overrightarrow {AC'}  = \frac{1}{2}\overrightarrow {AC}  \Rightarrow {V_{\left( {A;\frac{1}{2}} \right)}}\left( C \right) = C'\\{V_{\left( {A;\frac{1}{2}} \right)}}\left( A \right) = A.\\ \Rightarrow {V_{\left( {A;\frac{1}{2}} \right)}}\left( {ABC} \right) = AB'C'.\end{array}\)

Chọn C.

Gọi

\(\begin{array}{l}M'\left( {x;y} \right) = {V_{\left( {O;2} \right)}} \Leftrightarrow \overrightarrow {OM'}  = 2\overrightarrow {OM} \\ \Leftrightarrow \left( {x;y} \right) = 2\left( { - 3;5} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}x =  - 6\\y = 10\end{array} \right. \Rightarrow M'\left( { - 6;10} \right).\end{array}\)

Chọn C.