ΔBAC vuông tại B

=>\(BA^2+BC^2=AC^2\)

=>\(AC^2=\left(3a\right)^2+\left(4a\right)^2=9a^2+16a^2=25a^2=\left(5a\right)^2\)

=>AC=5a

\(\overrightarrow{MA}+\overrightarrow{MB}-\overrightarrow{MC}=\overrightarrow{0}\)

=>\(\overrightarrow{MB}+\overrightarrow{MA}+\overrightarrow{CM}=\overrightarrow{0}\)

=>\(\overrightarrow{MB}+\overrightarrow{CA}=\overrightarrow{0}\)

=>\(\overrightarrow{MB}=-\overrightarrow{CA}=\overrightarrow{AC}\)

=>AMBC là hình bình hành

Xét ΔABC vuông tại B có sin BAC=\(\frac{BC}{CA}=\frac{4a}{5a}=\frac45\) ; cos BAC=\(\frac{BA}{AC}=\frac{3a}{5a}=\frac35\)

\(\sin CAM=\sin\left(CAB+BAM\right)\)

\(=\sin90^0\cdot cosBAC+cos90^0\cdot\sin BAC=cosBAC\) =3/5

=>\(cosCAM=-\sqrt{1-\sin^2CAM}=-\frac45\)

N là trung điểm của AC

=>\(AN=\frac{AC}{2}=2,5a\)

AMBC là hình bình hành

=>AM=BC=4a

Xét ΔAMN có \(cosNAM=\frac{AN^2+AM^2-MN^2}{2\cdot AN\cdot AM}\)

=>\(\frac{\left(2.5a\right)^2+\left(4a\right)^2-MN^2}{2\cdot2,5a\cdot4a}=\frac{-4}{5}\)

=>\(6,25a^2+16a^2-MN^2=-\frac45\cdot5a\cdot4a=-4a\cdot4a=-16a^2\)

=>\(MN^2=6.25a^2+16a^2+16a^2=38,25a^2\)

=>\(MN=\sqrt{38,25a^2}=a\cdot\sqrt{\frac{153}{4}}=a\cdot\frac{3\sqrt{17}}{2}\)

Lời giải:

$|\overrightarrow{BC}|=BC=\sqrt{AB^2+AC^2}=\sqrt{(3a)^2+(4a)^2}=5a$ theo định lý Pitago.

Xíu nữa làm :v

1) Ta có:\(\overrightarrow{AB}+\overrightarrow{DE}-\overrightarrow{DB}+\overrightarrow{BC}=\overrightarrow{AE}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{BE}+\overrightarrow{EC}\)

\(=\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CE}+\overrightarrow{EC}=\overrightarrow{AC}+\overrightarrow{BE}\left(đpcm\right)\)2) a) Ta có: \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)

\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\left(đpcm\right)\)

b) Ta có: \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}\)

\(=\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{DB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{CB}\left(đpcm\right)\)c) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}-\overrightarrow{BD}\)

\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}\)

Ta có: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}+\overrightarrow{BC}\) ( đề bài bị lỗi gì à ?? :v ) hay do mình =))