b) 

1: Tọa độ A là:

\(\left\{{}\begin{matrix}x=0\\y=\left(m+1\right)\cdot x+3=0\left(m+1\right)+3=3\end{matrix}\right.\)

Vậy: A(0;3)

2: Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\\left(m+1\right)x+3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\x\left(m+1\right)=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{m+1}\end{matrix}\right.\)

=>\(B\left(\dfrac{-3}{m+1};0\right)\)

\(OB=\sqrt{\left(-\dfrac{3}{m+1}-0\right)^2+\left(0-0\right)^2}=\dfrac{3}{\left|m+1\right|}\)

\(OA=\sqrt{\left(0-0\right)^2+\left(3-0\right)^2}=3\)

OA=2OB

=>\(3=\dfrac{6}{\left|m+1\right|}\)

=>|m+1|=2

=>\(\left[{}\begin{matrix}m+1=2\\m+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-3\end{matrix}\right.\)

1: Thay x=0 và y=4 vào (d), ta được:

\(0\left(m^2+1\right)+m+2=4\)

=>m+2=4

=>m=2

2: tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\x\left(m^2+1\right)+m+2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{-m-2}{m^2+1}\\y=0\end{matrix}\right.\)

Tọa độ B là: \(\left\{{}\begin{matrix}x=0\\y=0\left(m^2+1\right)+m+2=m+2\end{matrix}\right.\)

vậy: O(0;0); \(A\left(\dfrac{-m-2}{m^2+1};0\right);B\left(0;m+2\right)\)

\(OA=\sqrt{\left(\dfrac{-m-2}{m^2+1}-0\right)^2+\left(0-0\right)^2}=\sqrt{\dfrac{\left(m+2\right)}{m^2+1}}^2=\dfrac{\left|m+2\right|}{m^2+1}\)

\(OB=\sqrt{\left(0-0\right)^2+\left(m+2-0\right)^2}=\sqrt{0^2+\left(m+2\right)^2}=\left|m+2\right|\)

Vì Ox\(\perp\)Oy nên ΔOAB vuông tại O

=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot\dfrac{\left(m+2\right)^2}{m^2+1}\)

Để \(S_{OBA}=\dfrac{1}{2}\) thì \(\dfrac{1}{2}\cdot\dfrac{\left(m+2\right)^2}{m^2+1}=\dfrac{1}{2}\)

=>\(\dfrac{\left(m+2\right)^2}{m^2+1}=1\)

=>\(\left(m+2\right)^2=m^2+1\)

=>\(m^2+4m+4=m^2+1\)

=>4m+4=1

=>4m=-3

=>\(m=-\dfrac{3}{4}\)

a:

b: phương trình hoành độ giao điểm là:

4x+2=2x-2

=>4x-2x=-2-2

=>2x=-4

=>x=-2

Thay x=-2 vào y=4x+2, ta được:

\(y=4\cdot\left(-2\right)+2=-8+2=-6\)

Vậy: M(-2;-6)

c: Tọa độ A là:

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

Tọa độ B là:

\(\left\{{}\begin{matrix}y=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\)

Vậy: B(1;0); A(-1/2;0)

d: M(-2;-6); B(1;0); A(-1/2;0)

\(MA=\sqrt{\left(-\dfrac{1}{2}+2\right)^2+\left(0-6\right)^2}=\dfrac{3\sqrt{17}}{2}\)

\(MB=\sqrt{\left(1+2\right)^2+\left(0+6\right)^2}=3\sqrt{5}\)

\(AB=\sqrt{\left(-\dfrac{1}{2}-1\right)^2+\left(0-0\right)^2}=\dfrac{3}{2}\)

Chu vi tam giác MAB là:

\(C_{MAB}=MA+MB+AB=\dfrac{3}{2}+3\sqrt{5}+\dfrac{3\sqrt{17}}{2}\)

Xét ΔMAB có \(cosAMB=\dfrac{MA^2+MB^2-AB^2}{2\cdot MA\cdot MB}=\dfrac{9}{\sqrt{85}}\)

=>\(sinAMB=\sqrt{1-\left(\dfrac{9}{\sqrt{85}}\right)^2}=\dfrac{2}{\sqrt{85}}\)

Diện tích tam giác MAB là:

\(S_{AMB}=\dfrac{1}{2}\cdot MA\cdot MB\cdot sinAMB=\dfrac{1}{2}\cdot\dfrac{3\sqrt{17}}{2}\cdot3\sqrt{5}\cdot\dfrac{2}{\sqrt{85}}\)

\(=\dfrac{9}{2}\)