\(x^2-\left(m+3\right)x+2m+2=0\Leftrightarrow\left(x-2\right)\left(x-m-1\right)=0\left(1\right)\)

\(\Delta=\left(m+3\right)^2-4\left(2m+2\right)=m^2+6m+9-8m-8=m^2-2m+1=\left(m-1\right)^2\ge0\left(\forall m\right)\)

\(\Rightarrow\left(1\right)\) \(luôn\) \(có\) \(nghiệm\) \(\forall m\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x=2\in(-\text{∞};3]\\x=m+1\end{matrix}\right.\)

\(\left(1\right)\) \(có\) \(đúng\) \(1\) \(nghiệm\) \(\in(-\text{∞};3]\)  \(\Leftrightarrow\left[{}\begin{matrix}x=m+1=2\\x=m+1>3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m=1\\m>2\end{matrix}\right.\)

\(\Rightarrow\left\{1\right\}\cup\left(2;+\text{∞}\right)\)

\(\)

B(2;-1) 

không vì \(2\cdot3^3>3-1\)

\(\overrightarrow{AH}=\left(x_H-1;y_H-2\right)\)

\(\overrightarrow{BC}=\left(11;2\right)\)

\(\overrightarrow{BH}=\left(x_H+2;y_H-6\right)\)

Theo đề, ta có:

\(\left\{{}\begin{matrix}11x_H-11+2y_H-4=0\\2x_H+4-11y_H+66=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}22x_H+4y_H=30\\22x_H-121y_H=-770\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}125y_H=800\\11x_H+2y_H=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y_H=\dfrac{32}{5}\\x_H=\dfrac{1}{5}\end{matrix}\right.\)

cái khúc từ đề bài ta có xuống í, làm sao bạn ra được cái hệ phương trình đó vậy ạ? 

Đề sai rồi bạn

Trục đối xứng là 2
=> -b/2a = 2
=> a = -b/4 = - (-4)/4 = 1
P đi qua A(1;2)
=> 2 = 1.1^2 - 4.1 + c
=> c + 1 - 4 = 2
=> c = 5
=> y = x^2 - 4x + 5

\(\left\{{}\begin{matrix}x_{B'}-5=2\\y_{B'}+6=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_{B'}=7\\y_{B'}=-2\end{matrix}\right.\)