Từ giả thiết ta có:
\(\left(\overrightarrow{a}+2\overrightarrow{b}\right)\left(5\overrightarrow{a}-4\overrightarrow{b}\right)=0\)
\(\Leftrightarrow\overrightarrow{a}.5\overrightarrow{a}-\overrightarrow{a}.4\overrightarrow{b}+2\overrightarrow{b}.5\overrightarrow{a}-2\overrightarrow{b}.4\overrightarrow{b}=0\)
\(\Leftrightarrow5a^2+6\overrightarrow{a}.\overrightarrow{b}-8b^2=0\)
\(\Leftrightarrow\left(5\overrightarrow{a}-4\overrightarrow{b}\right)\left(\overrightarrow{a}+2\overrightarrow{b}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\overrightarrow{a}=\dfrac{4}{5}\overrightarrow{b}\\\overrightarrow{a}=-2\overrightarrow{b}\end{matrix}\right.\)
Nếu \(\overrightarrow{a}=\dfrac{4}{5}\overrightarrow{b}\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=0^o\)
Nếu \(\overrightarrow{a}=-2\overrightarrow{b}\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=180^o\)
Làm lại đây nha, nãy buồn ngủ nên làm hơi ngu.
Từ giả thiết ta có:
\(\left(\overrightarrow{a}+2\overrightarrow{b}\right)\left(5\overrightarrow{a}-4\overrightarrow{b}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\overrightarrow{a}=\dfrac{4}{5}\overrightarrow{b}\\\overrightarrow{a}=-2\overrightarrow{b}\end{matrix}\right.\)
Nếu \(\overrightarrow{a}=\dfrac{4}{5}\overrightarrow{b}\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=0^o\)
Nếu \(\overrightarrow{a}=-2\overrightarrow{b}\Rightarrow\left(\overrightarrow{a};\overrightarrow{b}\right)=180^o\)
