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Câu 1:
ĐKXĐ:...
\(pt\Leftrightarrow\left(x-2\right)\left(x+1\right)\sqrt{x+1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (t/m)
Câu 2:
\(\overrightarrow{AB}+\overrightarrow{BD}=m\overrightarrow{AB}+n\overrightarrow{AC}\)
Có \(\overrightarrow{DC}=2\overrightarrow{BD}\Rightarrow\overrightarrow{DB}+\overrightarrow{BC}=2\overrightarrow{BD}\Leftrightarrow3\overrightarrow{BD}=\overrightarrow{BC}\)
Có \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}\Rightarrow\overrightarrow{BD}=\frac{1}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AD}=\overrightarrow{AB}+\frac{1}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}=\frac{2}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=m\overrightarrow{AB}+n\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{2}{3}\\n=\frac{1}{3}\end{matrix}\right.\)
Bài này thực chất là đi phân tích vt AD thành vt AB và AC thôi
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\Leftrightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\Rightarrow\overrightarrow{KA}=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{KA}+\overrightarrow{AD}=\left(-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\right)+\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow m-n=-\frac{1}{12}\)
\(\overrightarrow{MB}=-2\overrightarrow{MC}\Leftrightarrow\overrightarrow{MB}=-2\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)
\(\Rightarrow3\overrightarrow{MB}=-2\overrightarrow{BC}\Rightarrow\overrightarrow{BM}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{3}\\n=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow mn=\frac{2}{9}\)
a) Có \(\overrightarrow{BC}^2=\left(\overrightarrow{AC}-\overrightarrow{AB}\right)^2=\overrightarrow{AC}^2+\overrightarrow{AB}^2-2\overrightarrow{AC}.\overrightarrow{AB}\)
Suy ra: \(\overrightarrow{AC}.\overrightarrow{AB}=\dfrac{\overrightarrow{AC^2}+\overrightarrow{AB}^2-\overrightarrow{BC}^2}{2}=\dfrac{8^2+6^2-11^2}{2}=-\dfrac{21}{2}\).
Do \(\overrightarrow{AC}.\overrightarrow{AB}< 0\) nên \(cos\widehat{BAC}< 0\) suy ra góc A là góc tù.
b) Từ câu a suy ra: \(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=-\dfrac{21}{2.6.8}=-\dfrac{7}{32}\).
Do N là trung điểm của AC nên \(AN=AC:2=8:2=4cm\).
\(\overrightarrow{AM}.\overrightarrow{AN}=AM.AN.cos\left(\overrightarrow{AM},\overrightarrow{AN}\right)\)
\(=2.4.cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=2.4.\dfrac{-7}{32}=-\dfrac{7}{4}\).
Bài 1 và Bài 2 tương tự nhau nên mk sẽ chỉ CM bài 1 thôi nha
Có \(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\overrightarrow{AB}+\overrightarrow{CD}=0\)
\(\Rightarrow\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}=0\)
\(\Leftrightarrow\overrightarrow{AD}+\overrightarrow{CB}=0\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\)
Bài 3:
Xét \(\Delta AIP\) theo quy tắc trung điểm có:
\(\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}}{2}\)
Làm tương tự vs các tam giác còn lại
\(\Rightarrow\overrightarrow{IB}=\frac{\overrightarrow{IN}+\overrightarrow{IC}}{2}\)
\(\Rightarrow\overrightarrow{IA}=\frac{\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
Cộng vế vs vế
\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\frac{\overrightarrow{IA}+\overrightarrow{IP}+\overrightarrow{IN}+\overrightarrow{IC}+\overrightarrow{IB}+\overrightarrow{IM}}{2}\)
\(\Leftrightarrow2\overrightarrow{IA}+2\overrightarrow{IB}+2\overrightarrow{IC}=\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\)
\(\Leftrightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IM}+\overrightarrow{IN}+\overrightarrow{IP}\left(đpcm\right)\)
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\)
\(\Leftrightarrow-\overrightarrow{AC}=-3\overrightarrow{AN}\Rightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}\) (do M là trung điểm AB)
\(\overrightarrow{AK}=\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{6}\end{matrix}\right.\)
Ta có \(\overrightarrow{AD}=\overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AC}+2\overrightarrow{CB}=\overrightarrow{AC}+2\left(\overrightarrow{CA}+\overrightarrow{AB}\right)=\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AB}\)
\(=2\overrightarrow{AB}-\overrightarrow{AC}\)
Vậy m = 2 , n = -1