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Câu 1:
1: \(\overrightarrow{OM}=\dfrac{\overrightarrow{OA}+\overrightarrow{OD}}{2}=\dfrac{\overrightarrow{BO}+\overrightarrow{OA}}{2}=\dfrac{\overrightarrow{BA}}{2}=-\dfrac{1}{2}\overrightarrow{AB}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
\(\overrightarrow{BD}=2\cdot\overrightarrow{BO}=-2\cdot\overrightarrow{OB}\)
nên y=-2
2: \(2\cdot\overrightarrow{ON}=\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{DO}+\overrightarrow{OC}=\overrightarrow{DC}=\overrightarrow{AB}\)
Vậy: Các vecto u thỏa mãn là vecto DC và vecto AB
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Câu 5:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(c5\) \(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\left(đpcm\right)\)
\(c4:\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{GG'}\)
\(\overrightarrow{G'A'}+\overrightarrow{G'B'}+\overrightarrow{G'C'}=\overrightarrow{GG'}\)\(\Leftrightarrow\overrightarrow{G'A}+\overrightarrow{AA'}+\overrightarrow{G'B}+\overrightarrow{B'B}+\overrightarrow{G'C}+\overrightarrow{CC'}=\overrightarrow{GG'}\)
\(\Leftrightarrow\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\overrightarrow{GG'}\)\(\left(dpcm\right)\)
\(c3:a,\) \(2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IA}+2\overrightarrow{IM}=2\overrightarrow{MI}+2\left(\overrightarrow{IM}+\overrightarrow{MI}\right)=2.\overrightarrow{0}=\overrightarrow{0}\left(đpcm\right)\)
\(b,2\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)
\(=2\left(\overrightarrow{OI}+\overrightarrow{IA}\right)+\overrightarrow{OI}+\overrightarrow{IB}+\overrightarrow{OI}+\overrightarrow{IC}\)
\(=4\overrightarrow{OI}+2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\)\(=4\overrightarrow{OI}+\overrightarrow{0}=4\overrightarrow{OI}\left(đpcm\right)\)
\(c2:\) \(\left\{{}\begin{matrix}3AH=2AB\\3AK=AC\\4BM=3MC\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{AH}\\\overrightarrow{AC}=3\overrightarrow{AK}\\\overrightarrow{BM}=\dfrac{3}{7}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\dfrac{7}{3}\overrightarrow{BM}\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\)
\(\Rightarrow\dfrac{7}{3}\overrightarrow{BM}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\Leftrightarrow\overrightarrow{BM}=\dfrac{9}{7}\overrightarrow{AK}-\dfrac{9}{14}\overrightarrow{AH}\)
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11:
a: \(BD=AC=\sqrt{\left(3a\right)^2+\left(4a\right)^2}=5a\)
|vecto AB+vecto AD|
=|vecto AB+vecto BC|
=|vecto AC|
=5a
b: Gọi M là trung điểm của BC
=>BM=CM=BC/2=2a
\(AM=\sqrt{AB^2+BM^2}=a\sqrt{13}\)
Xét ΔABC có AM là trung tuyến
nên vecto AB+vecto AC=2*vecto AM
=>|vecto AB+vecto AC|=2|vecto AM|
=>\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=2\cdot AM=2a\sqrt{13}\)
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12:
a: Gọi M là trung điểm của BC
trên tia đối của tia MA, lấy D sao cho M là trung điểm của AD
Xét tứ giác ABDC có
M là trung điểm chung của AD và BC
góc BAC=90 độ
=>ABDC là hình chữ nhật
=>\(\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}\)
=>\(\overrightarrow{v}=\overrightarrow{AD}\)
b: \(\left|\overrightarrow{v}\right|=\left|\overrightarrow{AD}\right|=AD=2\cdot AM=2\cdot\dfrac{BC}{2}=BC=\sqrt{\left(6a\right)^2+\left(8a\right)^2}=10a\)
hình như đó đâu phải là câu 12 đâu ạ 🤔🤔🤔 trong đề câu 12 có chi tiết nào liên quan tới M đâu 🤔
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22:
a:
\(\overrightarrow{AD}=2\overrightarrow{DB}\)
=>\(\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\)
\(\overrightarrow{CE}=3\overrightarrow{EA}\)
=>\(\overrightarrow{AE}=\dfrac{1}{3}\overrightarrow{EC}\)
=>\(\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
Xét ΔAED có AM là trung tuyến
nên \(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)
b: \(\overrightarrow{MI}=\overrightarrow{ME}+\overrightarrow{EI}\)
\(=\dfrac{1}{2}\overrightarrow{DE}+\overrightarrow{EC}+\overrightarrow{CI}\)
\(=\dfrac{1}{2}\left(\overrightarrow{DA}+\overrightarrow{AE}\right)+\dfrac{3}{4}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)+\dfrac{3}{4}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{AB}\)
\(=\dfrac{-1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}+\dfrac{3}{4}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AB}\)
\(=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
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1.
\(x^3-mx^2+\left(m-2\right)x+1=0\)
\(\Leftrightarrow x^3-2x+1-m\left(x^2-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)-mx\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-\left(m-1\right)x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2-\left(m-1\right)x-1=0\left(1\right)\end{matrix}\right.\)
Để pt có 3 nghiệm pb \(\Leftrightarrow f\left(x\right)=x^2-\left(m-1\right)x-1=0\) có 2 nghiệm pb khác 1
\(\Leftrightarrow f\left(1\right)=1-\left(m-1\right)-1\ne0\) (pt trên hiển nhiên luôn có 2 nghiệm pb trái dấu do \(ac=-1< 0\))
\(\Leftrightarrow m\ne1\)
2.
\(x^3+\left(m+1\right)x^2+2mx+4=0\)
\(\Leftrightarrow x^3+x^2+4+mx\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+2\right)+mx\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+\left(m-1\right)x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x^2+\left(m-1\right)x+2=0\left(1\right)\end{matrix}\right.\)
Pt có 2 nghiệm khi:
TH1: (1) có nghiệm kép khác 2
\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(m-1\right)^2-8=0\\-\dfrac{b}{2a}=\dfrac{1-m}{2}\ne2\end{matrix}\right.\) \(\Rightarrow m=1\pm2\sqrt{2}\)
TH2: (1) có 2 nghiệm pb và 1 nghiệm trong đó bằng 2
\(\Rightarrow\left\{{}\begin{matrix}\Delta=\left(m-1\right)^2-8>0\\f\left(2\right)=4+2\left(m-1\right)+2=0\end{matrix}\right.\) \(\Rightarrow m=-2\)
Câu 1:
1: \(\overrightarrow{OM}=\dfrac{\overrightarrow{OA}+\overrightarrow{OD}}{2}=\dfrac{\overrightarrow{BO}+\overrightarrow{OA}}{2}=\dfrac{\overrightarrow{BA}}{2}=-\dfrac{1}{2}\overrightarrow{AB}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
\(\overrightarrow{BD}=2\cdot\overrightarrow{BO}=-2\cdot\overrightarrow{OB}\)
nên y=-2
2: \(2\cdot\overrightarrow{ON}=\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{DO}+\overrightarrow{OC}=\overrightarrow{DC}=\overrightarrow{AB}\)
Vậy: Các vecto u thỏa mãn là vecto DC và vecto AB