A

C

giải thích ?

a: A={\(x\in N\)|\(\left\{{}\begin{matrix}x< 20\\x⋮5\end{matrix}\right.\)}

b: B={\(x\in N\)|\(\left\{{}\begin{matrix}x=4k+1\\k\in N\end{matrix}\right.\)}

a.

Áp dụng định lý hàm cos:

\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=\sqrt{19}\)

Áp dụng công thức trung tuyến:

\(AM=\sqrt{\dfrac{2\left(AB^2+AC^2\right)-BC^2}{4}}=\dfrac{\sqrt{7}}{2}\)

b.

\(2\overrightarrow{IA}+\overrightarrow{IB}=\overrightarrow{0}\Leftrightarrow2\overrightarrow{IB}+2\overrightarrow{BA}+\overrightarrow{IB}=\overrightarrow{0}\Leftrightarrow\overrightarrow{IB}=\dfrac{2}{3}\overrightarrow{AB}\Rightarrow\overrightarrow{BI}=-\dfrac{2}{3}\overrightarrow{AB}\)

\(\overrightarrow{JB}-2\overrightarrow{JC}=\overrightarrow{0}\Rightarrow\overrightarrow{JB}=2\overrightarrow{JC}=2\overrightarrow{JB}+2\overrightarrow{BC}\Rightarrow\overrightarrow{JB}=-2\overrightarrow{BC}\Rightarrow\overrightarrow{BJ}=2\overrightarrow{BC}\)

\(\Rightarrow\overrightarrow{BI}.\overrightarrow{BJ}=-\dfrac{2}{3}\overrightarrow{AB}.2\overrightarrow{BC}=-\dfrac{4}{3}\overrightarrow{AB}.\overrightarrow{BC}=-\dfrac{4}{3}\overrightarrow{AB}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\dfrac{4}{3}AB^2-\dfrac{4}{3}\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{4}{3}.2^2-\dfrac{4}{3}.2.3.cosA=\dfrac{28}{3}\)

Độ dài IJ:

Ta có: \(\overrightarrow{BI}=-\dfrac{2}{3}\overrightarrow{AB}\Rightarrow BI=\dfrac{2}{3}AB=\dfrac{4}{3}\)

\(\overrightarrow{BJ}=2\overrightarrow{BC}\Rightarrow BJ=2BC=2\sqrt{19}\)

Từ đó:

\(IJ^2=\overrightarrow{IJ}^2=\left(\overrightarrow{IB}+\overrightarrow{BJ}\right)^2=IB^2+BJ^2+2\overrightarrow{IB}.\overrightarrow{BJ}\)

\(=IB^2+BJ^2-2\overrightarrow{BI}.\overrightarrow{BJ}\)

\(=\left(\dfrac{4}{3}\right)^2+\left(2\sqrt{19}\right)^2-2.\dfrac{28}{3}=...\)

Chọn A

Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(x+1;y-2\right)\\\overrightarrow{AB}=\left(-4;2\right)\\\overrightarrow{CH}=\left(x-2;y-4\right)\end{matrix}\right.\)

CH là đường cao hạ từ C \(\Rightarrow\left\{{}\begin{matrix}H\in AB\\CH\perp AB\end{matrix}\right.\)

 \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{-4}=\dfrac{y-2}{2}\\-4\left(x-2\right)+2\left(y-4\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}\\y=\dfrac{6}{5}\end{matrix}\right.\)

\(\Rightarrow H\left(\dfrac{3}{5};\dfrac{6}{5}\right)\)

Chọn C

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