b: Tọa độ của F là:

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

\(a,m=-\dfrac{3}{2}\Leftrightarrow x^2-2\cdot\dfrac{1}{2}\cdot x-\dfrac{3}{2}+1=0\\ \Leftrightarrow x^2-x-\dfrac{1}{2}=0\\ \Leftrightarrow2x^2-2x-1=0\\ \Delta'=1+2=3\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{3}}{2}\\x=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\\ b,\text{PT có }n_o\Leftrightarrow\Delta'=\left(m+2\right)^2-\left(m+1\right)\ge0\\ \Leftrightarrow m^2+3m+3\ge0\\ \Leftrightarrow\left(m+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge0\left(\text{luôn đúng}\right)\)

Vậy PT có nghiệm với mọi m

\(c,\text{Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)\\x_1x_2=m+1\end{matrix}\right.\\ x_1\left(1-2x_2\right)+x_2\left(1-2x_1\right)=m^3\\ \Leftrightarrow\left(x_1+x_2\right)-4x_1x_2=m^3\\ \Leftrightarrow2\left(m+2\right)-4\left(m+1\right)=m^3\\ \Leftrightarrow m^3+2m=0\\ \Leftrightarrow m\left(m^2+2\right)=0\Leftrightarrow m=0\)

a: Xét ΔABD có 

\(BD^2=AD^2+AB^2\)

nên ΔABD vuông tại A

\(a,m=1\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\\dfrac{x}{2}-\dfrac{y}{3}=334\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\\dfrac{1+y}{2}-\dfrac{y}{3}=334\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1+y\\3y+3-2y=2004\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1+y\\y=2001\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2002\\y=2001\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}mx-y=1\\\dfrac{x}{2}-\dfrac{y}{3}=334\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx-y=1\\3x-2y=2004\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}mx-y=1\\y=\dfrac{3x-2004}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}mx-\dfrac{3x-2004}{2}=1\\y=\dfrac{3x-2004}{2}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2mx-3x=-2002\\y=\dfrac{3x-2004}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(2m-3\right)=-2002\\y=\dfrac{3x-2004}{2}\end{matrix}\right.\)

Để hpt vô nghiệm thì \(x\left(2m-3\right)=-2002\) vô nghiệm

\(\Leftrightarrow2m-3=0\Leftrightarrow m=\dfrac{3}{2}\)

( Hình em tự vẽ nhé )

+ Ta có: ΔABC = ΔDEF

=> \(\widehat{A}=\widehat{D}=30^o\)

+ Ta có: \(2\widehat{B}=3\widehat{C}\)

=> \(\widehat{B}=\dfrac{3\widehat{C}}{2}\)

+ Xét ΔABC 

=> \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(t3g\Delta\right)\)

Mà \(\widehat{A}=30^o;\widehat{B}=\dfrac{3\widehat{C}}{2}\)

=> \(30^o+\dfrac{3\widehat{C}}{2}+\widehat{C}=180^o\)

=> \(\dfrac{3\widehat{C}}{2}+\widehat{C}=150^o\)

\(\Rightarrow\dfrac{3\widehat{C}}{2}+\dfrac{2\widehat{C}}{2}=150^o\)

\(\Rightarrow\dfrac{5\widehat{C}}{2}=150^o\)

\(\Rightarrow5\widehat{C}=75^o\)

\(\Rightarrow\widehat{C}=15^o\)

+ Xét ΔABC

\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(t3g\Delta\right)\)

\(\Rightarrow30^o+15^o+\widehat{B}=180^o\)

\(\Rightarrow\widehat{B}=135^o\)

Bài 1:

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

hay \(\widehat{BOC}=135^0\)

Tách ra đi bạn nhé!

Bài 2: 

Tọa độ giao điểm là:

\(\left\{{}\begin{matrix}2x+y=3\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+2y=6\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Thay x=1 và y=1 vào (d), ta được:

\(2m-1+1=5m\)

hay m=0