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\(\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NC}+\overrightarrow{BM}+\overrightarrow{MN}+\overrightarrow{ND}\)
\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{BM}\right)+\left(\overrightarrow{NC}+\overrightarrow{ND}\right)\)
\(=2\overrightarrow{MN}\)
\(\Rightarrow k=\dfrac{1}{2}\)
\(\overrightarrow{AB}+\overrightarrow{B_1C_1}+\overrightarrow{DD_1}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CC_1}=\overrightarrow{AC_1}\)
\(\Rightarrow k=1\)
\(\dfrac{MA}{MB}=k\Rightarrow MA=kMB=k\left(AB-AM\right)\Rightarrow MA=\dfrac{k}{k+1}AB\)
\(\Rightarrow\overrightarrow{MA}=\dfrac{k}{k+1}\overrightarrow{BA}\)
Tương tự: \(\overrightarrow{CN}=\dfrac{k}{k+1}\overrightarrow{CD}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AC}+\overrightarrow{CN}=\dfrac{k}{k+1}\overrightarrow{BA}+\overrightarrow{AC}+\dfrac{k}{k+1}\overrightarrow{CD}\)
\(=\dfrac{k}{k+1}\left(\overrightarrow{BD}+\overrightarrow{DA}\right)+\overrightarrow{AC}+\dfrac{k}{k+1}\overrightarrow{CD}\)
\(=\dfrac{k}{k+1}\overrightarrow{BD}+\dfrac{k}{k+1}\left(\overrightarrow{CD}+\overrightarrow{DA}\right)+\overrightarrow{AC}\)
\(=\dfrac{k}{k+1}\overrightarrow{BD}-\dfrac{k}{k+1}\overrightarrow{AC}+\overrightarrow{AC}\)
\(=\dfrac{k}{k+1}\overrightarrow{BD}+\dfrac{1}{k+1}\overrightarrow{AC}\)
\(\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AM}+\overrightarrow{MN}+\overrightarrow{NC}+\overrightarrow{BM}+\overrightarrow{MN}+\overrightarrow{ND}\)
\(=2\overrightarrow{MN}+\left(\overrightarrow{AM}+\overrightarrow{BM}\right)+\left(\overrightarrow{NC}+\overrightarrow{ND}\right)\)
\(=2\overrightarrow{MN}\)
\(\Rightarrow\) A đúng nên D sai
D. k=\(\dfrac{1}{2}\)