......................?

mik ko biết

mong bn thông cảm 

nha ................

\(\left(m+1\right)x+\left(m-2\right)y=3\)\(\left(m\ne-1;m\ne2\right)\)

\(y=0\Leftrightarrow x=\dfrac{3}{m+1}\Rightarrow A\left(\dfrac{3}{m+1};0\right)\Rightarrow OA=\left|\dfrac{3}{m+1}\right|\)

\(x=0\Leftrightarrow y=\dfrac{3}{m-2}\Leftrightarrow B\left(0;\dfrac{3}{m-2}\right)\Rightarrow OB=\left|\dfrac{3}{m-2}\right|\)

\(S_{_{ }^{ }\Delta ABO}=\dfrac{9}{2}=\dfrac{1}{2}OA.OB=\dfrac{1}{2}.\dfrac{9}{\left|m+1\right|.\left|m-2\right|}\Leftrightarrow\dfrac{1}{\left|m+1\right|.\left|m-2\right|}=9\Leftrightarrow\left|m+1\right|.\left|m-2\right|=9\Leftrightarrow\left(m+1\right)^2.\left(m-2\right)^2-81=0\Leftrightarrow\left(m^2-m-11\right)\left(m^2-m+7\right)=0\Leftrightarrow\left[{}\begin{matrix}m^2-m-11=0\Leftrightarrow m=\dfrac{1\pm3\sqrt{5}}{2}\left(tm\right)\\m^2-m+7=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow m=\dfrac{1\pm3\sqrt{5}}{2}\)

Cho x = 0 => \(y=\dfrac{3}{m-2}\)

vậy d cắt Oy tại A(0;3/m-2) => Oy = \(\left|\dfrac{3}{m-2}\right|\)

Cho y = 0 => \(x=\dfrac{3}{m+1}\)

vậy d cắt Ox tại B(3/m+1;0) => Ox = \(\left|\dfrac{3}{m+1}\right|\)

Ta có : \(S_{OAB}=\dfrac{1}{2}.OB.OA=\dfrac{1}{2}.\dfrac{9}{\left|\left(m+1\right)\left(m-2\right)\right|}=\dfrac{9}{2}\)

\(\Leftrightarrow\left|\left(m+1\right)\left(m-2\right)\right|=1\Leftrightarrow\left[{}\begin{matrix}m^2-m-3=0\\m^2-m-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1+\sqrt{13}}{2};m=\dfrac{1-\sqrt{13}}{2}\\m=\dfrac{1+\sqrt{5}}{2};m=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

Gọi A,B lần lượt là giao điểm của (d) với trục Ox và Oy

Tọa độ A là:

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

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

=>\(A\left(\dfrac{2}{2m+1};0\right)\)

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

Tọa độ B là:

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

=>B(0;-2)

\(OB=\sqrt{\left(0-0\right)^2+\left(-2-0\right)^2}=\sqrt{0+4}=2\)

Vì Ox\(\perp\)Oy

nên OA\(\perp\)OB

=>ΔOAB vuông tại O

=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{1}{2}\cdot2\cdot\dfrac{2}{\left|2m+1\right|}=\dfrac{2}{\left|2m+1\right|}\)

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

=>|2m+1|=2

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

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

a: Để (d)//(d') thì m+2=-2

hay m=-4