bai 1: cho tam giac ABC co trung tuyen BM =6, CN =9, goc BGC = 120 do( G la trong tam cua tam giac ABC). tinh do dai cac canh tam giac ABC

bai2: a/ xac dinh diem M tren elip: x^{2} + 4y^{2} = 4 sao cho khoang cach tu M den duong thang denta co phuong trinh: 3x + 4y - 2007 = 0 dat gia tri lon nhat
        b/ xac dinh diem M thuoc E : \frac{x^{2}}{18} + \frac{y^{2}}{8} = 1 sao cho khoang cach tu M den duong thang d: 2x - 3y + 25 = 0 dat gia tri nho nhat
Bài 2 . Khoảng cách từ $M(x,y)$ đến  $\triangle$
$d=\frac{|3x+4y-2007|}{\sqrt{3^2+4^2}}=\frac{|3x+4y-2007|}{5}$
Áp dụng BĐT Bunhia
$(x^2+4y^2)(9+4)\ge (3x+4y)^2$
$\Rightarrow (3x+4y)^2\le 4.13$
$\Rightarrow -2\sqrt{13}\le 3x+4y\le 2\sqrt{13}$
$\Rightarrow |3x+4y-2007|\le 2007+2\sqrt{13}$
Dấu bằng có $\Leftrightarrow 3x+4y=-2\sqrt{13}$
$\Leftrightarrow x= -\frac{6}{\sqrt{13}} , y=-\frac{2}{\sqrt{13}}$
Vậy khoảng cách lớn nhất là
$d=\frac{2007+2\sqrt{13}}{5}$  khi  $M(-\frac{6}{\sqrt{13}},-\frac{2}{\sqrt{13}}$

Cách làm tương tự cho câu b
Bài 1.
$\triangle BGC$ có  $BG=4, CG=6 , \widehat{BGC}=120$
$BC^2=BG^2+CG^2-BG.CG.2cos120=4^2+6^2+4.6=76$
$\Rightarrow BC=\sqrt{76}=2\sqrt{19}$

$\triangle MGC$ có  $MG=2 , CG=6 , \widehat{MGC}=60$
$\Rightarrow MC^2=MG^2+CG^2-MG.CG.2cos60=2^2+6^2-2.6=28$
$\Rightarrow AC=2MC=4\sqrt7$

$\triangle NGB$  có $NG=3, BG=4 , \widehat{NGB}=60$
$\Rightarrow NB^2=NG^2+BG^2-NG.BG.2cos60=3^2+4^2-3.4=13$
$\Rightarrow AB=2NB=2\sqrt{13}$


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