Giải thích các bước giải:
Ta có $c^2=2(ac+bc-ab)$
$\to c^2-2ac-2bc+2ab=0$
Ta có:
$\dfrac{a^2+(a-c)^2}{b^2+(b-c)^2}$
$=\dfrac{a^2+c^2-2ac-2bc+2ab+(a-c)^2}{b^2+c^2-2ac-2bc+2ab+(b-c)^2}$
$=\dfrac{(a-c)^2-2bc+2ab+(a-c)^2}{(b-c)^2-2ac+2ab+(b-c)^2}$
$=\dfrac{2(a-c)^2-2bc+2ab}{2(b-c)^2-2ac+2ab}$
$=\dfrac{(a-c)^2-bc+ab}{(b-c)^2-ac+ab}$
$=\dfrac{(a-c)^2+b(a-c)}{(b-c)^2+a(b-c)}$
$=\dfrac{(a-c)(a-c+b)}{(b-c)(b-c+a)}$
$=\dfrac{(a-c)(a-c+b)}{(b-c)(a-c+b)}$
$=\dfrac{a-c}{b-c}$
