câu hình;

a) Xét \(\Delta BDC\) và \(\Delta BAC:\) Ta có: \(\left\{{}\begin{matrix}BD=BA\\BCchung\\CD=CA\end{matrix}\right.\)

\(\Rightarrow\Delta BDC=\Delta BAC\left(c-c-c\right)\)

b) \(\Delta BDC=\Delta BAC\Rightarrow\angle BDC=\angle BAC=90\Rightarrow ABDC\) nội tiếp

c) \(\) MD cắt (C,CA) tại N'

Ta có: \(\angle AMN'=\angle AMD=\dfrac{1}{2}\angle ABD=\angle ABC\)

Lại có: \(\angle AN'M=\angle AN'D=\dfrac{1}{2}\angle ACD=\angle ACB\)

mà \(\angle ABC+\angle ACB=90\Rightarrow\angle AMN'+\angle AN'M=90\Rightarrow\angle MAN'=90\)

\(\Rightarrow AM\bot AN'\Rightarrow N\equiv N'\Rightarrow\) đpcm

câu đồ thị:

b) pt hoành độ giao điểm: \(\dfrac{x^2}{2}-\left(m-2\right)x-m+1=0\)

\(\Rightarrow x^2-2\left(m-2\right)x-2m+2=0\)

\(\Delta'=\left(m-2\right)^2+2m-2=m^2-2m+2=\left(m-1\right)^2+1>0\)

\(\Rightarrow\) pt luôn có 2 nghiệm phân biệt

Ta có : 

\(\left(x+2\right)^2=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\) 

\(\left(x+2\right)^2=0\)

\(< =>x+2=0\)

\(< =>x=-2\)

Ta có : \(\left(x+2\right)^2\ge0\forall x\)

Dấu = xảy ra <=> x + 2 = 0

                     <=> x = -2

Vậy với x = -2 thì \(\left(x+2\right)^2=0\)

(3x)^2=3^2.x^2=9x^2

Tham khảo nhé!

I want to raise money for the street children because I think it's the best way to help them. I will ask for the help from everybody to raise a fund. With this fund, we can provide the street children the basic needs for life. Then, I want to train them. I want them to have a job so that they can have a better life. They won't have to wander on the street and no one can treat them badly. I will ask my friends for help to realize this idea.

Lời giải:
$-4.2(-4+2)+(-4-2)^2=-8(-2)+(-6)^2=16+36=52$