Giải thích các bước giải:
Ta có : $M\in d\to M(a,a)$
$\to MA+MB=\sqrt{(a-3)^2+(a-2)^2}+\sqrt{(a-5)^2+(a-1)^2}$
$\to MA+MB=\sqrt{2a^2-10a+13}+\sqrt{2a^2-12a+26}$
$\to MA+MB=\sqrt{2\left(a-\dfrac{5}{2}\right)^2+\dfrac{1}{2}}+\sqrt{2(a-3)^2+8}$
$\to MA+MB=\sqrt{\left(a\sqrt{2}-\dfrac{5}{\sqrt{2}}\right)^2+\dfrac{1}{2}}+\sqrt{(a\sqrt{2}-3\sqrt{2})^2+8}$
$\to MA+MB=\sqrt{\left(a\sqrt{2}-\dfrac{5}{\sqrt{2}}\right)^2+\dfrac{1}{2}}+\sqrt{(3\sqrt{2}-a\sqrt{2})^2+8}$
$\to MA+MB=\sqrt{\left(a\sqrt{2}-\dfrac{5}{\sqrt{2}}+3\sqrt{2}-a\sqrt{2}\right)^2+(\sqrt{\dfrac{1}{2}}+2\sqrt{2})^2}$
$\to MA+MB=\sqrt{13}$
Dấu = xảy ra khi :
$\dfrac{a\sqrt{2}-\dfrac{5}{\sqrt{2}}}{3\sqrt{2}-a\sqrt{2}}=\dfrac{\sqrt{\dfrac{1}{2}}}{2\sqrt{2}}$
$\to a=\dfrac{13}5\to M(\dfrac{13}5,\dfrac{13}5)$
