\(A=\left(-3;-1\right)\cup\left(1;2\right)\)

\(B=\left(-1;+\infty\right)\)

\(C=\left(-\infty;2m\right)\)

\(A\cap B=\left(-3;-1\right)\)

Để \(A\cap B\cap C\ne\varnothing\Leftrightarrow2m\ge-1\)

\(\Leftrightarrow m\ge-\dfrac{1}{2}\)

Vậy \(m\ge-\dfrac{1}{2}\) thỏa đề bài

Lời giải:

$A\cap B\cap C=A\cap (B\cap C)$

Để tập hợp trên khác rỗng thì trước hết $B\cap C\neq \varnothing$

Điều này xảy ra khi $2m>m\Leftrightarrow m>0$

Khi đó: $B\cap C=(m; 2m)$

$\Rightarrow A\cap B\cap C=((-3;-1)\cup (1;2))\cap (m; 2m)$

$=((-3;-1)\cap (m;2m))\cup ((1;2)\cap (m; 2m))$

$=(1;2)\cap (m; 2m)$ (do $m>0$)

Để $(1;2)\cap (m; 2m)\neq \varnothing$ thì:

\(\left\{\begin{matrix} 2m>1\\ m< 2\end{matrix}\right.\Leftrightarrow m\in (\frac{1}{2};2)\)

Vậy...........

\(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)

\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)=\left(-2;7\right)\)

\(\left(0;12\right)\cap[5;+\infty)=[5;12)\)

\(R\cap\left[-1;1\right]=\left[-1;1\right]\)

\(A\cup B=[-2;+\infty)\)

\(\Rightarrow C_R\left(A\cup B\right)=\left(-\infty;-2\right)\)

\(mx^2-4x+m-3=0\left(1\right)\)

Để tập hợp B có đúng 2 tập con và \(B\subset A\) thì \(\left(1\right)\) có 2 nghiệm phân biệt cùng dương

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\P>0\\S>0\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}4-m\left(m-3\right)>0\\\dfrac{m-3}{m}>0\\\dfrac{4}{m}>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-3m-4< 0\\m< 0\cup m>3\\m>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1< m< 4\\m< 0\cup m>3\\m>0\end{matrix}\right.\)

\(\Leftrightarrow3< m< 4\)

Ta có:

\(\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{BG}\) 

+) \(\overrightarrow{BG}=\dfrac{1}{3}\left(\overrightarrow{BM}+\overrightarrow{BN}\right)=\dfrac{1}{3}\left(-\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CN}\right)\)

          \(=\dfrac{1}{3}\left(-\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{AC}-\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{DC}\right)=\dfrac{1}{3}\left(-\dfrac{13}{6}\overrightarrow{AB}+\overrightarrow{AC}\right)\)

          \(=-\dfrac{13}{18}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)

=> \(\overrightarrow{AG}=\dfrac{5}{18}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)

Mặt khác:

\(\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}=\overrightarrow{AB}+k\overrightarrow{BC}=\overrightarrow{AB}+k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)

Để A, G, I thẳng hàng 

=>\(\dfrac{\dfrac{5}{18}}{1-k}=\dfrac{\dfrac{1}{3}}{k}\Rightarrow k=\dfrac{6}{11}\)

Dễ thấy nếu \(A\cap B=\varnothing\Rightarrow A\in[-3;3)\Rightarrow\left\{{}\begin{matrix}m-1\ge-3\\\dfrac{m+3}{2}< 3\end{matrix}\right.\)

                                                               \(\Leftrightarrow-2\le m< 3\)

Do đó để \(A\cap B\ne\varnothing\Rightarrow m\notin[-2;3)\Rightarrow\left[{}\begin{matrix}m< -2\\m\ge3\end{matrix}\right.\)

A\(\cup\)B=(-∞;3)

A\(\cap\)B=(-5;-1\(]\)

B\(/\)A=(-1;3)

B=