​Từ giả thiết, ta suy ra tam giác $ABC$ có $\widehat{CAB}={60}^{\circ }$, $\widehat{ABC}={105}^{\circ }30^{\prime }$ và $AB=70.$

Khi đó $\hat{A}+\hat{B}+\hat{C}={180}^{\circ }\iff \hat{C}={180}^{\circ }-\left(\hat{A}+\hat{B}\right)={180}^{\circ }-{165}^{\circ }30^{\prime }={14}^{\circ }30^{\prime }$.

Theo định lí sin, ta có $\genfrac{}{}{0ex}{}{AC}{\sin B}=\genfrac{}{}{0ex}{}{AB}{\sin C}$ hay $\genfrac{}{}{0ex}{}{AC}{\sin {105}^{\circ }30^{\prime }}=\genfrac{}{}{0ex}{}{70}{\sin {14}^{\circ }30^{\prime }}$.

Do đó $AC=\genfrac{}{}{0ex}{}{70.\sin {105}^{\circ }30^{\prime }}{\sin {14}^{\circ }30^{\prime }}\approx 269,4$ m.

Gọi $CH$ là khoảng cách từ $C$ đến mặt đất. Tam giác vuông $ACH$ có cạnh $CH$ đối diện với góc ${30}^{\circ }$ nên $CH=\genfrac{}{}{0ex}{}{AC}{2}=\genfrac{}{}{0ex}{}{269,4}{2}=134,7$ m.

Vậy ngọn núi cao khoảng $134,7$ m.

Ta có $\overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OA}-\overrightarrow{DA}=\overrightarrow{OA}+\overrightarrow{AD}=\overrightarrow{OD}$

$\Rightarrow \left|\overrightarrow{OA}-\overrightarrow{CB}\right|=\left|\overrightarrow{OD}\right|=OD=\genfrac{}{}{0ex}{}{1}{2}BD=\genfrac{}{}{0ex}{}{1}{2}\sqrt{A{B}^2+A{D}^2}=\genfrac{}{}{0ex}{}{1}{2}\sqrt{a^2+a^2}=\genfrac{}{}{0ex}{}{a\sqrt{2}}{2}$

b) $VT=\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}=\left(\overrightarrow{DA}-\overrightarrow{DB}\right)+\overrightarrow{DC}=\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{CD}+\overrightarrow{DC}=\overrightarrow{0}=VP$

x−2y​

2x+y+1​<=>3(x−2y)>2(2x+y+1)

<=>3x−6y>4x+2y+2

<=>4x−3x+2y+6y<−2

<=>x+8y<−2

a,{2,3,4}

{0,1,2,3,4,5,6}

b.txd: D=(-∞;2]