Bài 1:
Để A giao B bằng rỗng thì \(\left[{}\begin{matrix}m+3< -3\\2m-1>6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m< -6\\m>\dfrac{7}{2}\end{matrix}\right.\)

1.

\(A\subset B\Leftrightarrow\left\{{}\begin{matrix}2m-1\le-1\\2m+3\ge1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\le0\\m\ge-1\end{matrix}\right.\) \(\Rightarrow-1\le m\le0\)

\(B\subset A\Leftrightarrow\left\{{}\begin{matrix}-1\le2m-1\\2m+3\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge0\\m\le-1\end{matrix}\right.\) \(\Rightarrow\) ko tồn tại m thỏa mãn yêu cầu

\(A\cap B\) nhưng bằng cái gì? Chỗ này đề thiếu

2.

a.

\(B\subset A\Leftrightarrow\left\{{}\begin{matrix}-4\le m-7\\m\le3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge3\\m\le3\end{matrix}\right.\) \(\Leftrightarrow m=3\)

b.

\(A\cup B=A\Leftrightarrow B\subset A\Leftrightarrow\left\{{}\begin{matrix}m\ge-3\\m\le1\\-4\le-3\end{matrix}\right.\) \(\Rightarrow-3\le m\le1\)

c.

\(A\backslash B=\varnothing\Leftrightarrow A\subset B\Leftrightarrow\left\{{}\begin{matrix}m-1< 5\\m-1\ge3\end{matrix}\right.\) \(\Rightarrow4\le m< 6\)

1.Nếu như có số tự nhiên k (kEN)sao cho (a +b) = m.k

2.________________________________(a - b)______

3_________________________________(a + b + c) = m.k

Ta có:

1+\(\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)

Thay a=1

=>\(1+\dfrac{1}{b}=b+\dfrac{1}{c}=c+1\)

*Lấy \(1+\dfrac{1}{b}=c+1\Rightarrow\dfrac{1}{b}=c\Rightarrow b=\dfrac{1}{c}\)

=>\(1+\dfrac{1}{b}=\dfrac{2}{c}=c+1\)

*Lấy \(\dfrac{2}{c}=\dfrac{c+1}{1}\)

=> 2=c(c+1)

<=> 2=c²+c

=>c=-2

*Lấy \(1+\dfrac{1}{b}=\dfrac{2}{c}\)

Thay c=-2 và quy đồng

=>\(\dfrac{b+1}{b}=-1\)

=>b+1=-b

=> b+b=-1

=>2b=-1

=> b=-1/2

Vậy b=\(-\dfrac{1}{2};c=-2\)

câu 3b) 0