Tọa độ giao điểm A,B là nghiệm của hệ phương trình:

\(\left\{{}\begin{matrix}x^2=2x+3\\y=2x+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+1\right)=0\\y=2x+3\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in\left\{\left(3;9\right);\left(-1;1\right)\right\}\)

vậy: A(3;9); B(-1;1)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\sqrt{5}-1\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(-8-2\sqrt{5}\right)y=6\sqrt{5}+2\\2x-3y=3\sqrt{5}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-\sqrt{5}\\x=\dfrac{3\sqrt{2}-3\sqrt{5}+2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\3.\left(\sqrt{5}-1\right)x+6y=3-3\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(3\sqrt{5}+1\right)x=1+3\sqrt{5}\\y=\dfrac{3\sqrt{5}-1-2x}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\sqrt{5}-1-2.1}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{-3.\left(1-\sqrt{5}\right)}{-3}=1-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left(x;y\right)=\left(1;1-\sqrt{5}\right)\)

a) Tứ giác ABCD là hình vuông (gt).

\(\Rightarrow\widehat{BCD}=90^o00\) (Tính chất hình vuông).

Xét tứ giác DBHC:

\(\widehat{BCD}=\widehat{BHD}\left(=90^o\right).\)

Mà 2 đỉnh H; C kề nhau cùng nhìn cạnh BD.

\(\Rightarrow\) Tứ giác DBHC nội tiếp (dhnb).

b) Xét \(\Delta HKD\) và \(\Delta CKB:\)

\(\widehat{K}chung.\)

\(\widehat{DHK}=\widehat{BCK}\left(=90^o\right).\)

\(\Rightarrow\text{​​}\Delta HKD\sim\Delta CKB\left(g-g\right).\)

\(\Rightarrow\dfrac{KH}{KC}=\dfrac{KD}{KB}\) (2 cạnh tương ứng tỉ lệ).

\(\Rightarrow KC.KD=KH.KB.\)

Bn tk:

\(\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

a: \(x_1+x_2=3;x_1x_2=-7\)

\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=3^2-2\cdot\left(-7\right)=9+14=23\)

b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{3}{-7}=\dfrac{-3}{7}\)

c: \(=\sqrt{\left(x_1-x_2\right)^2}=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{3^2-4\cdot\left(-7\right)}=\sqrt{37}\)

a, Xet tam giac ABC vuong tai A, duong cao AH

* Ap dung he thuc : \(AH^2=BH.CH\Rightarrow AH=\sqrt{16}=4cm\)

b, Do AB = AC => tam giac ABC can tai A

=> AH la duong cao dong thoi la duong phan giac 

Xet tam giac AMH va tam giac ANH co 

^AMH = ^ANH = 90⁰

^MAH = ^NAH ( AH la duong phan giac ) 

AH _ chung 

Vay tam giac AMH = tam giac ANH ( ch - gn )

=> AM = AN (1) ; Lai co : BA - AM = MB ; AC - AN = NC

=> MB = NC (2) 

Lay (1) . (2) ta duoc AM . MB = AN . CN 

(d) đi qua A(2;1) <=> 2a + b = 1 (1) 

(d) đi qua B(-1;0) <=> -a + b = 0 (2) 

Từ (1) ; (2) ta có hệ \(\left\{{}\begin{matrix}2a+b=1\\-a+b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ptđt (d) có dạng y = 1/3x + 1/3 

\(\left(x+2y\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+2y=0\\y-1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)

Bn tk câu a và c nha: