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