Đáp án:
$\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}=2012$
Giải thích các bước giải:
Ta có:
$\quad \dfrac{a^2}{a+b} +\dfrac{b^2}{b+c} +\dfrac{c^2}{c+a}=2012$
$\to \dfrac{a^2- b^2 + b^2}{a+b} +\dfrac{b^2- c^2 + c^2}{b+c} +\dfrac{c^2- a^2 + a^2}{c+a}=2012$
$\to \dfrac{a^2 - b^2}{a+b} +\dfrac{b^2}{a+b}+\dfrac{b^2-c^2}{b+c} +\dfrac{c^2}{b+c}+\dfrac{c^2-a^2}{c+a}+\dfrac{a^2}{c+a}=2012$
$\to \dfrac{(a-b)(a+b)}{a+b} +\dfrac{(b-c)(b+c)}{b+c} +\dfrac{(c-a)(c+a)}{c+a} + \left(\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}\right)=2012$
$\to a-b + b-c + c - a + \left(\dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}\right)=2012$
$\to \dfrac{b^2}{a+b} +\dfrac{c^2}{b+c} +\dfrac{a^2}{c+a}=2012$
