C16:
Giả sử tồn tại bộ số (k;h) sao cho \(\overrightarrow{c}=k\cdot\overrightarrow{a}+h\cdot\overrightarrow{b}\)
\(\rightarrow\left\{{}\begin{matrix}5=2\cdot k+1\cdot h\\0=-2\cdot k+4\cdot h\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}k=2\\h=1\end{matrix}\right.\)
Vậy \(\overrightarrow{c}=2\overrightarrow{a}+\overrightarrow{b}\)
C17:
Gọi tọa độ điểm M thuộc Ox là (a;0)
Ta có: \(\overrightarrow{MA}=\left(a-1;0-0\right)=\left(a-1;0\right)\)
\(\overrightarrow{MB}=\left(a-0;0-3\right)=\left(a;-3\right)\)
\(\overrightarrow{MC}=\left(a-\left(-3\right);0-\left(-5\right)\right)=\left(a+3;5\right)\)
\(2\overrightarrow{MA}-3\overrightarrow{MB}+2\overrightarrow{MC}=\left\{{}\begin{matrix}x=2\left(a-1\right)-3a+2\left(a+3\right)\\y=2\cdot0-3\cdot\left(-3\right)+2\cdot5\end{matrix}\right.=\left\{{}\begin{matrix}x=a+4\\y=19\end{matrix}\right.=\left(a+4;19\right)\)\(T=\left|2\overrightarrow{MA}-3\overrightarrow{MB}+2\overrightarrow{MC}\right|=\sqrt{\left(a+4\right)^2+19^2}=\sqrt{\left(a+4\right)^2+361}\)
Ta có: \(\left(a+4\right)^2\ge0\rightarrow\left(a+4\right)^2+361\ge361\rightarrow\sqrt{\left(a+4\right)^2+361}\ge\sqrt{361}=19\)
\(T_{min}=19\Leftrightarrow\left(a+4\right)^2=0\Leftrightarrow a=-4\)
Vậy M có tọa độ (-4;0)
C18:
\(\overrightarrow{u}\) cùng phương với \(\overrightarrow{v}\Leftrightarrow\overrightarrow{u}=k\cdot\overrightarrow{v}\left(k\in R\right)\)
\(\rightarrow\left\{{}\begin{matrix}2x-1=1\cdot k\\3=k\cdot\left(x+2\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}2x-1=k\\\left(2x-1\right)\left(x+2\right)=3\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}2x-1=k\\2x^2+3x-2=3\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}2x-1=k\\2x^2+3x-5=0\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}2x-1=k\\\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
Tích hai giá trị của x là \(-\dfrac{5}{2}\cdot1=-\dfrac{5}{2}\)
