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1.
Dựng \(\overrightarrow{DB'}=\overrightarrow{CB}\)
\(k\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{DB}\)
\(=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DA}+\overrightarrow{AB}\)
\(=2\overrightarrow{AB}+\overrightarrow{B'D}+\overrightarrow{DA}\)
\(=2\overrightarrow{AB}+\overrightarrow{B'A}\)
\(=2\overrightarrow{AB}+2\overrightarrow{AB}=4\overrightarrow{AB}\)
\(\Rightarrow k=4\)
Gọi M là trung điểm IB
\(\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\left|2\overrightarrow{AM}\right|=2AM\)
Ta có \(\overrightarrow{AM}^2=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2=MI^2+IA^2-2MI.IA.cos90^o=\dfrac{1}{16}a^2+\dfrac{3}{4}a^2=\dfrac{13}{16}a^2\)
\(\Rightarrow AM=\dfrac{\sqrt{13}}{4}a\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AI}\right|=\dfrac{\sqrt{13}}{2}a\)
\(\overrightarrow{MN}=\overrightarrow{MC}+\overrightarrow{CN}=\dfrac{3}{4}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}=\dfrac{3}{4}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)-\dfrac{1}{2}\overrightarrow{AB}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\)
\(\Rightarrow a+b=\dfrac{1}{2}+\dfrac{3}{4}=...\)
\(\left|\overrightarrow{MA}+\overrightarrow{MC}-\overrightarrow{MN}\right|=\left|\overrightarrow{MA}+\overrightarrow{MD}+\overrightarrow{DC}-\overrightarrow{MN}\right|\)\(=\left|\overrightarrow{DC}-\frac{1}{2}\overrightarrow{DC}-\frac{1}{2}\overrightarrow{AB}\right|=\left|\overrightarrow{DC}-\frac{3}{4}\overrightarrow{DC}\right|=\frac{1}{A}DC=\frac{a}{2}\)