Giải thích các bước giải:
Đặt $P=A\cdot B\cdot C$
$\to P=\dfrac{4bc+a^2}{2a^2-bc}\cdot \dfrac{4ca-b^2}{ca+2b^2}\cdot \dfrac{4ab+c^2}{2c^2-ab}$
Vì $a-b+c=0$
$\to b=a+c$
$\to P=\dfrac{4(a+c)c+a^2}{2a^2-(a+c)c}\cdot \dfrac{4ca-(a+c)^2}{ca+2(a+c)^2}\cdot \dfrac{4a(a+c)+c^2}{2c^2-a(a+c)}$
$\to P=\dfrac{a^2+4ac+4c^2}{2a^2-ac-c^2}\cdot \dfrac{4ca-a^2-2ac-c^2}{ac+2a^2+4ac+2c^2}\cdot \dfrac{4a^2+4ac+c^2}{2c^2-a^2-ac}$
$\to P=\dfrac{(a+2c)^2}{(a-c)(2a+c)}\cdot \dfrac{-(a-c)^2}{(2a+c)(a+2c)}\cdot \dfrac{(2a+c)^2}{(c-a)(2c+a)}$
$\to P=1$
$\to đpcm$
