Giải thích các bước giải:
2.Từ câu $1b\to (d_1)\cap (d_2)=A(0,3)$
Ta có $(d_1)\cap Ox$ tại $B(-\dfrac{3}{m},0)$
$(d_2)\cap Ox$ tại $C(3m,0)$
Ta có:
$m\cdot \dfrac{-1}m=-1\to (d_1)\perp (d_2)$
$\to AB\perp AC$
$\to S_{ABC}=\dfrac12\cdot AB\cdot AC$
$\to S_{ABC}=\dfrac12\cdot \sqrt{(-\dfrac3m-0)^2+(0-3)^2}\cdot \sqrt{(3m-0)^2+(0-3)^2}$
$\to S_{ABC}=\dfrac12\cdot \sqrt{\dfrac{9}{m^2}+9}\cdot \sqrt{9m^2+9}$
$\to S_{ABC}=\dfrac12\cdot \sqrt{(\dfrac{9}{m^2}+9)\cdot (9m^2+9)}$
$\to S_{ABC}=\dfrac92\cdot \sqrt{(\dfrac{1}{m^2}+1)\cdot (m^2+1)}$
$\to S_{ABC}=\dfrac92\cdot \sqrt{\dfrac{m^2+1}{m^2}\cdot (m^2+1)}$
$\to S_{ABC}=\dfrac92\cdot \sqrt{\dfrac{(m^2+1)^2}{m^2}}$
$\to S_{ABC}\ge \dfrac92\cdot \sqrt{\dfrac{4m^2}{m^2}}$
$\to S_{ABC}\ge \dfrac92\cdot \sqrt{4}$
$\to S_{ABC}\ge 9$
Dấu = xảy ra khi $m^2=1\to m=\pm1$
