1:

(d): y=mx-2x+1=x(m-2)+1

Để (d)//(d') thì

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

=>\(\left\{{}\begin{matrix}m=0\\m< >\dfrac{1}{3}\end{matrix}\right.\)

=>m=0

2: Để \(sinBAO=\dfrac{\sqrt{5}}{5}\) thì góc tạo bởi (d) với trục Ox có sin bằng \(\dfrac{\sqrt{5}}{5}\)

\(cosBAO=\sqrt{1-sin^2BAO}=\sqrt{1-\dfrac{1}{5}}=\dfrac{2}{\sqrt{5}}\)

\(tanBAO=\dfrac{sinBAO}{cosBAO}=\dfrac{1}{\sqrt{5}}:\dfrac{2}{\sqrt{5}}=\dfrac{1}{2}\)

=>\(a=tanBAO=\dfrac{1}{2}\)

=>m-2=1/2

=>m=5/2

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

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

1: Bạn bổ sung đề bài đi bạn

2: Tọa độ A là:

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

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

=>\(OA=\sqrt{\left(\dfrac{4}{2m-1}-0\right)^2+\left(0-0\right)^2}=\dfrac{4}{\left|2m-1\right|}\)

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=\left(2m-1\right)x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\left(2m-1\right)\cdot0-4=-4\end{matrix}\right.\)

=>OB=4

Để ΔOAB cân tại O thì OA=OB

=>\(\dfrac{4}{\left|2m-1\right|}=4\)

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

=>\(\left|2m-1\right|=1\)

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

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

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

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

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

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

a) \(y=\left(1-m\right)x+m+2\left(d\right)\)

\(y=2x-1\left(d'\right)\)

\(\left(d\right)//\left(d'\right)\Leftrightarrow\left\{{}\begin{matrix}1-m=2\\m+2\ne-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=-1\\m\ne-3\end{matrix}\right.\)

\(\Leftrightarrow m=-1\)

Vậy với \(m=-1\) để \(\left(d\right)//\left(d'\right)\)

b) \(\left(d\right)\cap\left(Ox\right)=A\left(x;0\right)\)

\(\Leftrightarrow\left(1-m\right)x+m+2=0\)

\(\Leftrightarrow x=\dfrac{m-1}{m+2}\)

\(\Rightarrow A\left(\dfrac{m-1}{m+2};0\right)\)

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

\(\left(d\right)\cap\left(Oy\right)=B\left(0;y\right)\)

\(\Leftrightarrow\left(1-m\right).0+m+2=y\)

\(\Leftrightarrow y=m+2\)

\(\Rightarrow B\left(0;m+2\right)\)

\(\Rightarrow OB=\sqrt[]{\left(m+2\right)^2}=\left|m+2\right|\)

Để \(\Delta OAB\) là \(\Delta\) vuông cân khi và chỉ khi

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}m^2+2m+4=m-1\\m^2+2m+4=1-m\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m^2+m+5=0\left(1\right)\\m^2+3m+3=0\left(2\right)\end{matrix}\right.\)

Giải \(pt\left(1\right):\Delta=1-20=-19< 0\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Giải \(pt\left(2\right):\Delta=9-12=-3< 0\)

\(\Rightarrow\left(2\right)\) vô nghiệm

Vậy không có giá trị nào của \(m\) thỏa mãn đề bài

a: PTHĐGĐ là:

x^2-(2m+1)x+2m-4=0

Δ=(2m+1)^2-4(2m-4)

=4m^2+4m+1-8m+16

=4m^2-4m+17=(2m-1)^2+16>=16>0 với mọi m

=>(P) luôn cắt (d) tại hai điểm phân biệt

b: x1+x2=2m+1;x1x2=2m-4

HK=4

=>|x1-x2|=4

=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=4\)

=>\(\sqrt{4m^2+4m+1-8m+16}=4\)

=>4m^2-4m+17=16

=>m=1/2