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mik 2k7 nhung co
bai j khong biet
ban se gui cho mik nhe!
mik lam cho
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3.
\(\overrightarrow{IA}+3\overrightarrow{IC}=\overrightarrow{0}\Rightarrow\overrightarrow{IB}+\overrightarrow{BA}+3\overrightarrow{IB}+3\overrightarrow{BC}=\overrightarrow{0}\)
\(\Rightarrow4\overrightarrow{IB}+\overrightarrow{BA}+3\overrightarrow{BC}=\overrightarrow{0}\Rightarrow4\overrightarrow{IB}=\overrightarrow{AB}+3\overrightarrow{CB}\) (1)
\(\overrightarrow{JA}+2\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\Rightarrow\overrightarrow{JB}+\overrightarrow{BA}+2\overrightarrow{JB}+3\overrightarrow{JB}+3\overrightarrow{BC}=\overrightarrow{0}\)
\(\Rightarrow6\overrightarrow{JB}+\overrightarrow{BA}+3\overrightarrow{BC}=0\Rightarrow6\overrightarrow{JB}=\overrightarrow{AB}+3\overrightarrow{CB}\) (2)
(1);(2) \(\Rightarrow4\overrightarrow{IB}=6\overrightarrow{JB}\Rightarrow\overrightarrow{IB}\) và \(\overrightarrow{JB}\) cùng phương
Hay I; J; B thẳng hàng
4.
\(\overrightarrow{PA}+\overrightarrow{PB}=\overrightarrow{0}\Rightarrow\overrightarrow{PA}+\overrightarrow{PA}+\overrightarrow{AB}=0\Rightarrow\overrightarrow{PA}=-\dfrac{1}{2}\overrightarrow{AB}\)
\(\overrightarrow{NA}=3\overrightarrow{CN}\Rightarrow\overrightarrow{NA}=3\overrightarrow{CA}+3\overrightarrow{AN}\Rightarrow4\overrightarrow{AN}=3\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AN}=\dfrac{3}{4}\overrightarrow{AC}\)
\(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)
\(\Rightarrow2\overrightarrow{BM}=3\overrightarrow{BC}\Rightarrow\overrightarrow{BM}=\dfrac{3}{2}\overrightarrow{BC}=\dfrac{3}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\dfrac{3}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{PN}=\overrightarrow{PA}+\overrightarrow{AN}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\) (1)
\(\overrightarrow{PM}=\overrightarrow{PB}+\overrightarrow{BM}=\dfrac{1}{2}\overrightarrow{AB}-\dfrac{3}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}=-\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}=2\left(-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)\) (2)
(1);(2) \(\Rightarrow\overrightarrow{PM}=2\overrightarrow{PN}\Rightarrow\) P, M, N thẳng hàng
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Do \(\overrightarrow{MN}=\overrightarrow{DA}\Rightarrow ANMD\) là hình bình hành
Theo giả thiết: \(\overrightarrow{AM}=\overrightarrow{BA}\Leftrightarrow\overrightarrow{MA}=\overrightarrow{AB}\)
Mà \(\overrightarrow{AB}=\overrightarrow{DC}\) do ABCD là hbh
\(\Rightarrow\overrightarrow{MA}=\overrightarrow{DC}\)
Lại có: \(\overrightarrow{MN}=\overrightarrow{DA}\Leftrightarrow\overrightarrow{NM}=\overrightarrow{AD}\)
Do đó:
\(\overrightarrow{NA}=\overrightarrow{NM}+\overrightarrow{MA}=\overrightarrow{AD}+\overrightarrow{DC}=\overrightarrow{AC}\) (đpcm)
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a: (2x-1)(x+3)<=0
nên -3<=x<=1/2
b: \(\left(2x-7\right)\left(4-5x\right)>=0\)
=>(2x-7)(5x-4)<=0
=>4/5<=x<=7/2
i: \(\Leftrightarrow\dfrac{3-x+2}{x-2}>0\)
\(\Leftrightarrow\dfrac{x-5}{x-2}< 0\)
=>2<x<5
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a.
Do (P) qua M và N nên:
\(\left\{{}\begin{matrix}a-b+4=7\\16a-4b+4=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-4\end{matrix}\right.\)
b.
Do (P) có trục đối xứng x=2 và qua A nên:
\(\left\{{}\begin{matrix}-\dfrac{b}{2a}=2\\4a+2b+4=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4a+b=0\\4a+2b=-8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=-8\end{matrix}\right.\)
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lấy bán kính chia cho độ dài cung sẽ ra được số đo radian
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