Giải thích các bước giải:
Ta có:
$A=\sqrt{x^2+4y^2+6x+9}+\sqrt{x^2+4y^2-2x-12y+10}$
$\to A=\sqrt{(x+3)^2+4y^2}+\sqrt{(x-1)^2+(2y-3)^2}$
$\to A=\sqrt{(x+3)^2+(2y)^2}+\sqrt{(1-x)^2+(3-2y)^2}$
$\to A\ge \sqrt{(x+3+1-x)^2+(2y+3-2y)^2}$
$\to A\ge 5$
Dấu = xảy ra khi:
$\dfrac{x+3}{1-x}=\dfrac{2y}{3-2y}$
$\to \dfrac{x+3}{1-x}+1=\dfrac{2y}{3-2y}+1$
$\to \dfrac{x+3+1-x}{1-x}=\dfrac{2y+3-2y}{3-2y}$
$\to \dfrac{4}{1-x}=\dfrac{3}{3-2y}$
$\to 3(1-x)=4(3-2y)$
$\to 3x-8y+9=0$
