Cho $y$ $=$$-2x^{2}$
 Cho A = $\left ( \frac{-2}{3} ;-7\right )$ B =$\left ( 2;1 \right )$
a )Viết phương trình đường thẳng AB (d)
b ) Xác định tọa độ giao điểm của (P) và (d) 
Tìm điểm trên (P) 
 sao cho tổng hoành độ và tung độ = -6
Cách 9 
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=> \frac{x+\frac{2}{3}}{2+\frac{2}{3}}=\frac{y+7}{1+7}$
=>$8x+\frac{16}{3}-\frac{8}{3}y-\frac{56}{3}=0=>y=3x-5$
chưa hok ak –  Yêu Tatoo 15-04-17 11:28 PM
chưa............................ –  ๖ۣۜTQT☾♋☽ 15-04-17 09:05 AM
học r mà . =.= –  Ryo 15-04-17 09:04 AM
hình như lp 9 nó chưa hok cái ct đầu đâu,phải cm ra –  ๖ۣۜTQT☾♋☽ 15-04-17 09:02 AM
$1) \overrightarrow{AB}=(\frac{8}{3},8)=>\overrightarrow{n}=(-8,\frac{8}{3})$
$=>pt(d): -8(x+\frac{2}{3})+\frac{8}{3}(y+7)=0$
$y=3x-5$
2) pt hoành độ giao điểm : $-2x^2=3x-5=>x_1=1=>A(1,-2)$
$x_2=\frac{-5}{2}=>B(\frac{-5}{2},\frac{-25}{2})$
2*) theo đề ta có : $y_c+x_c=-6=>y_c=-6-x_c$
Vì C thuộc (P): $y_c=-2x_c^2=>-2x_c^2=-6-x_c=>x_{c1}=2$=>$C_1(2,-8)$
$y_{C2}=\frac{-3}{2}=>C_2=(\frac{-3}{2},\frac{-9}{2})$

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