Cho  $\overrightarrow{a}(2;1), \overrightarrow{b}(3;-4), \overrightarrow{c}(-7;2).   $
a) Tìm tọa độ của vec-tơ  $\overrightarrow{u}=3 \overrightarrow{a}+2 \overrightarrow{b}-4\overrightarrow{c};    $
b) Tìm tọa độ của vec-tơ  $\overrightarrow{x} $  sao cho  $\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}    $;
c) Tìm các số  $k,h$  sao cho  $\overrightarrow{c}=k \overrightarrow{a}+h \overrightarrow{b}.   $
a) Ta có:  $\overrightarrow{u}=(3.2+2.3-4(-7); 3.1+2.(-4)-4.2) $
$\Rightarrow    \overrightarrow{u}=(40;-13) $.

b) Gọi tọa độ của  $x$  là  $(m;n)$  ta có:  $\overrightarrow{x}+\overrightarrow{a}=(m+2; n+1)  $
                                                                      $\overrightarrow{b}-\overrightarrow{c}=(10;-6)  $
Giải hệ:   $\begin{cases}m+2=10 \\ n+1=-6 \end{cases}     \Rightarrow      m=8;  n=-7$
                                                     $\Rightarrow      \overrightarrow{x}=(8;-7) $

c) Ta có:  $\overrightarrow{c}=k \overrightarrow{a}+h \overrightarrow{b}    \Rightarrow    \overrightarrow{c}=(2k+3h;k-4h) $
Với  $\overrightarrow{c}=(-7; 2) $  cho ta hệ   $\begin{cases}2k+3h=-7 \\ k-4h=2 \end{cases} $
Giải hệ này ta được   $k=-2;  h=-1$
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