sr nha em ko làm dc

a: Tọa độ A là:

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

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

vậy: \(A\left(-\dfrac{3}{m+1};0\right)\)

Tọa độ B 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: B(0;3)

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

\(OB=\sqrt{\left(0-0\right)^2+\left(3-0\right)^2}=\sqrt{0+9}=3\)

Vì Ox\(\perp\)Oy

nên OA\(\perp\)OB

=>ΔOAB vuông tại O

=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot3\cdot\dfrac{3}{\left|m+1\right|}=\dfrac{9}{2\left|m+1\right|}\)

Để \(S_{AOB}=9\) thì \(\dfrac{9}{2\left|m+1\right|}=9\)

=>2|m+1|=1

=>|m+1|=1/2

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

Để ĐTHS cắt cả 2 trục tọa độ \(\Rightarrow m\ne0\)

Khi đó ta có: giao điểm với trục hoành: \(mx+2=0\Rightarrow x=-\dfrac{2}{m}\)

Giao điểm với trục tung: \(y=m.0+2=2\)

a. \(A\left(-\dfrac{2}{m};0\right)\Rightarrow OA=\left|x_A\right|=\left|\dfrac{2}{m}\right|\)

\(B\left(0;2\right)\Rightarrow OB=\left|y_B\right|=2\)

\(OA=OB\Rightarrow\left|\dfrac{2}{m}\right|=2\Rightarrow m=\pm1\)

b. \(C\left(-\dfrac{2}{m};0\right);D\left(0;2\right)\Rightarrow\left\{{}\begin{matrix}OC=\left|\dfrac{2}{m}\right|\\OD=2\end{matrix}\right.\)

\(tanC=\dfrac{OD}{OC}=\left|m\right|=2\Rightarrow m=\pm2\)

Bài 1:

a: Để hàm số y=(1-m)x+m+2 đồng biến trên R thì 1-m>0

=>-m>-1

=>m<1

b: Tọa độ A là:

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

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

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=\left(1-m\right)x+m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\left(1-m\right)\cdot0+m+2=m+2\end{matrix}\right.\)

=>\(OB=\left|m+2\right|\)

Để ΔOAB cân tại O thì OA=OB

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

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

=>\(\left[{}\begin{matrix}m+2=0\\\dfrac{1}{\left|m-1\right|}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-2\\m-1=1\\m-1=-1\end{matrix}\right.\)

=>\(m\in\left\{0;2;-2\right\}\)

Tọa độ A là;

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

Tọa độ B là:

x=0 và y=(m+1)*0+3=3

=>OB=3

S_OAB=9

=>1/2*OA*OB=9

=>1/2*9/|m+1|=9

=>1/2*1/|m+1|=1

=>1/|m+1|=2

=>|m+1|=1/2

=>m+1=1/2 hoặc m+1=-1/2

=>m=-1/2 hoặc m=-3/2