a: \(4\cdot\overrightarrow{AC}=3\cdot\overrightarrow{AB}+\overrightarrow{AD}\)
=>\(4\cdot\left(\overrightarrow{AD}+\overrightarrow{DC}\right)=3\left(\overrightarrow{AD}+\overrightarrow{DB}\right)+\overrightarrow{AD}\)
=>\(4\cdot\overrightarrow{AD}+4\cdot\overrightarrow{DC}=4\cdot\overrightarrow{AD}+3\cdot\overrightarrow{DB}\)
=>\(4\cdot\overrightarrow{DC}=3\cdot\overrightarrow{DB}\)
=>\(\overrightarrow{DC}=\frac34\cdot\overrightarrow{DB}\)
=>D,C,B thẳng hàng
b:
\(\overrightarrow{DC}=\frac34\cdot\overrightarrow{DB}\)
=>C nằm giữa D và B sao cho \(DC=\frac34DB\)
DC+CB=DB
=>CB=DB-DC=DB-3/4DB=1/4DB
=>DC=3CB
=>\(\overrightarrow{CD}+3\cdot\overrightarrow{CB}=\overrightarrow{0}\)
\(\overrightarrow{MD}+3\cdot\overrightarrow{MB}\)
\(=\overrightarrow{MC}+\overrightarrow{CD}+3\cdot\overrightarrow{MC}+3\cdot\overrightarrow{CB}\)
\(=4\cdot\overrightarrow{MC}+\left(\overrightarrow{CD}+3\cdot\overrightarrow{CB}\right)\)
\(=4\cdot\overrightarrow{MC}\)
