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A sai
\(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{DA}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{DB}=-\overrightarrow{BD}\) mới đúng
b.\(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\) là đẳng thức đúng
a, \(AC=\dfrac{AB}{sin45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}=a.a\sqrt{2}.cos45^o=a^2\)
b, \(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)
\(=\overrightarrow{AC}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BC}\)
\(=AC.BD.cos90^o+AC.AD.cos45^o\)
\(=a\sqrt{2}.a\sqrt{2}.0+a\sqrt{2}.a.\dfrac{\sqrt{2}}{2}=a^2\)
c, \(\overrightarrow{AB}.\overrightarrow{BD}=AB.BD.cos135^o=-a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2\)
d, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\left(2\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{BC}.\left(\overrightarrow{AD}+\overrightarrow{BD}\right)\)
\(=\overrightarrow{BC}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BD}\)
\(=AD^2+BC.BD.cos45^o\)
\(=a^2+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=2a^2\)
e, \(\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right)\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\right)\)
\(=\left(\overrightarrow{AC}+\overrightarrow{AC}\right)\left(\overrightarrow{DB}+\overrightarrow{DB}\right)\)
\(=4.\overrightarrow{AC}.\overrightarrow{DB}=4.AC.DB.cos90^o=0\)
a) Theo quy tắc hình bình hành ta có: \(\overrightarrow {AB} + \overrightarrow {AD} = \overrightarrow {AC} \)
\( \Rightarrow |\overrightarrow {AB} + \overrightarrow {AD} |\; = \;|\overrightarrow {AC} |\)
Vậy mệnh đề này đúng.
b) Ta có: \(\overrightarrow {AB} + \overrightarrow {BD} = \overrightarrow {AD} = \overrightarrow {BC} \ne \overrightarrow {CB} \)
Vậy mệnh đề này sai.
c) Ta có: \(\overrightarrow {OA} + \overrightarrow {OB} = \overrightarrow {OC} + \overrightarrow {OD} \)\( \Leftrightarrow \overrightarrow {OA} - \overrightarrow {OD} + \overrightarrow {OB} - \overrightarrow {OC} = \overrightarrow {0} \Leftrightarrow \overrightarrow {DA} + \overrightarrow {CB} =\overrightarrow {0}\Leftrightarrow 2\overrightarrow {CB} =\overrightarrow {0} \)
Vậy mệnh đề này sai.
\(a\text{) }\overrightarrow{AB}-\overrightarrow{CD}=\left(\overrightarrow{AC}+\overrightarrow{CB}\right)-\overrightarrow{CD}\\ =\overrightarrow{AC}-\left(\overrightarrow{CD}-\overrightarrow{CB}\right)=\overrightarrow{AC}-\overrightarrow{BD}\)
\(b\text{) }\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}=\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)\\ =\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)=\overrightarrow{AD}+\overrightarrow{DA}=0\)
\(c\text{) }\overrightarrow{AC}+\overrightarrow{DE}-\overrightarrow{DC}-\overrightarrow{CE}+\overrightarrow{CB}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DE}-\overrightarrow{DC}\right)-\overrightarrow{CE}\\ =\overrightarrow{AB}+\overrightarrow{CE}-\overrightarrow{CE}=\overrightarrow{AB}\)
\(d\text{) }\overrightarrow{AB}+\overrightarrow{DE}+\overrightarrow{CF}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DF}+\overrightarrow{FE}\right)+\left(\overrightarrow{CE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}+\left(\overrightarrow{FE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}\)
Câu 1: A
$\overrightarrow{BA}-\overrightarrow{BC}+\overrightarrow{DC}=\overrightarrow{CA}+\overrightarrow{DC}=\overrightarrow{DA}=\overrightarrow{CB}$
Câu 2:
$\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{DB}$
$=\overrightarrow{DC}+\overrightarrow{CB}$
$\Rightarrow \overrightarrow{AB}-\overrightarrow{DC}=\overrightarrow{CB}+\overrightarrow{AD}$
$\Rightarrow \overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}$
Đáp án A.
Tham khảo:
A. Ta có: \(\left( {\overrightarrow {AB} ,\overrightarrow {BD} } \right) = \left( {\overrightarrow {BE} ,\overrightarrow {BD} } \right) = {135^o} \ne {45^o}.\) Vậy A sai.
B. Ta có: \(\left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right) = \left( {\overrightarrow {CF} ,\overrightarrow {CG} } \right) = {45^o}\) và \(\overrightarrow {AC} .\overrightarrow {BC} = AC.BC.\cos {45^o} = a\sqrt 2 .a.\frac{{\sqrt 2 }}{2} = {a^2}.\)
Vậy B đúng.
Chọn B
C. Dễ thấy \(AC \bot BD\) nên \(\overrightarrow {AC} .\overrightarrow {BD} = 0 \ne {a^2}\sqrt 2.\) Vậy C sai.
D. Ta có: \(\left( {\overrightarrow {BA} .\overrightarrow {BD} } \right) = {45^o}\) \( \Rightarrow \overrightarrow {BA} .\overrightarrow {BD} = BA.BD.\cos {45^o} = a.a\sqrt 2 .\frac{{\sqrt 2 }}{2} = {a^2} \ne - {a^2}.\) Vậy D sai.
a) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {ND} \\= \left( {\overrightarrow {AM} + \overrightarrow {BM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) \\= \overrightarrow 0 + 2\overrightarrow {MN} + \overrightarrow 0 = 2\overrightarrow {MN} \) (đpcm)
b) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
\(\)\(\overrightarrow {BC} + \overrightarrow {AD} = \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {ND} \)
\(\left( {\overrightarrow {BM} + \overrightarrow {AM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) = 2\overrightarrow {MN} \)
Mặt khác ta có: \(\overrightarrow {AC} + \overrightarrow {BD} = 2\overrightarrow {MN} \)
Suy ra \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
Cách 2:
\(\begin{array}{l}
\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \\
\Leftrightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \\
\Leftrightarrow \overrightarrow {DC} = \overrightarrow {DC} (đpcm)
\end{array}\)
Ta có:
\(\overrightarrow {MN} = \overrightarrow {MA} + \overrightarrow {AD} + \overrightarrow {DN} \)
Mặt khác: \(\overrightarrow {MN} = \overrightarrow {MB} + \overrightarrow {BC} + \overrightarrow {CN} \)
\(\begin{array}{l} \Rightarrow 2\overrightarrow {MN} = \overrightarrow {MA} + \overrightarrow {AD} + \overrightarrow {DN} + \overrightarrow {MB} + \overrightarrow {BC} + \overrightarrow {CN} \\ \Leftrightarrow 2\overrightarrow {MN} = \left( {\overrightarrow {MA} + \overrightarrow {MB} } \right) + \left( {\overrightarrow {DN} + \overrightarrow {CN} } \right) + \overrightarrow {BC} + \overrightarrow {AD} \\ \Leftrightarrow 2\overrightarrow {MN} = \overrightarrow 0 + \overrightarrow 0 + \overrightarrow {BC} + \overrightarrow {AD} \\ \Leftrightarrow 2\overrightarrow {MN} = \overrightarrow {BC} + \overrightarrow {AD} \end{array}\)
Lại có:
\(\overrightarrow {BC} + \overrightarrow {AD} = \overrightarrow {BD} + \overrightarrow {DC} + \overrightarrow {AD} = \overrightarrow {AD} + \overrightarrow {DC} + \overrightarrow {BD} = \overrightarrow {AC} + \overrightarrow {BD} .\)
Vậy \(\overrightarrow {BC} + \overrightarrow {AD} = 2\overrightarrow {MN} = \;\overrightarrow {AC} + \overrightarrow {BD} .\)
Đẳng thức đúng là: \(\overrightarrow{AB}+\overrightarrow{BD}=2\overrightarrow{BC}\)
Vậy chọn câu a)