Giúp vớii mn
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giúp em vớii
NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 8 2021
a: \(\widehat{xOm};\widehat{nOm}\)
\(\widehat{nOm};\widehat{yOn}\)
\(\widehat{yOn};\widehat{xOn}\)
\(\widehat{xOm};\widehat{yOm}\)
b: \(\widehat{yOn};\widehat{xOn}\)
\(\widehat{xOm};\widehat{yOm}\)
c: \(\widehat{mOy}=180^0-30^0=150^0\)
\(\widehat{nOx}=180^0-50^0=130^0\)
\(\widehat{mOn}=180^0-80^0=100^0\)
13 tháng 12 2021
\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;-7\right)\\\overrightarrow{BC}=\left(4;8\right)\\\overrightarrow{AC}=\left(8;1\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{BC}+\overrightarrow{AC}=\left(12,9\right)\)
\(\Rightarrow\overrightarrow{AB}.\left(\overrightarrow{BC}+\overrightarrow{AC}\right)=4.12-7.9=...\)
b. Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(x+3;y-5\right)\\\overrightarrow{BH}=\left(x-1;y+2\right)\\\end{matrix}\right.\)
Do \(AH\perp BC\Rightarrow\overrightarrow{AH}.\overrightarrow{BC}=0\)
\(\Rightarrow4\left(x+3\right)+8\left(y-5\right)=0\)
\(\Rightarrow x+2y=7\) (1)
Do H thuộc BC \(\Rightarrow\dfrac{x-1}{4}=\dfrac{y+2}{8}\Rightarrow2x-y=4\Rightarrow y=2x-4\)
Thế vào (1) \(\Rightarrow x+2\left(2x-4\right)=7\Rightarrow x=3\Rightarrow y=2\)
\(\Rightarrow H\left(3;2\right)\)
a: Ta có: NA=2NC
=>\(S_{PNA}=2\times S_{PNC};S_{MNA}=2\times S_{MNC}\)
=>\(S_{PNA}-S_{MNA}=2\times\left(S_{PNC}-S_{MNC}\right)\)
=>\(S_{PMA}=2\times S_{PMC}\)
Ta có: MB=1/3MC
=>\(S_{PMB}=\frac13\times S_{PMC}\)
=>\(\frac{S_{PBM}}{S_{PAM}}=\frac13:2=\frac16\)
=>\(\frac{PB}{PA}=\frac16\)
=>\(\frac{PB}{BA}=\frac15\)
=>\(\frac{PB}{15}=\frac15\)
=>BP=3(cm)
AP=AB+BP
=3+15=18(cm)
b:
BM+MC=BC
=>BC=3MB+MB=4BM
=>\(BM=\frac14\times BC\)
=>\(S_{ABM}=\frac14\times S_{ABC}\)
Vì \(\frac{BP}{BA}=\frac{3}{15}=\frac15\)
nên \(S_{PBM}=\frac15\times S_{BAM}=\frac15\times\frac14\times S_{ABC}=\frac{1}{20}\times S_{ABC}\)
\(S_{PMA}=S_{ABM}+S_{PBM}=\frac14\times S_{ABC}+\frac{1}{20}\times S_{ABC}=\frac{3}{10}\times S_{ABC}\)
Vì \(\frac{CM}{CB}=\frac34\)
nên \(S_{AMC}=\frac34\times S_{ABC}\)
Ta có: AN+NC=AC
=>AC=2NC+NC=3NC
=>\(AN=\frac23AC\)
=>\(S_{AMN}=\frac23\times S_{AMC}=\frac23\times\frac34\times S_{ABC}=\frac12\times S_{ABC}\)
=>\(\frac{S_{AMP}}{S_{AMN}}=\frac{3}{10}:\frac12=\frac{3}{10}\times2=\frac35\)
=>\(\frac{MP}{MN}=\frac35\)