Đáp án: $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$
Giải thích các bước giải:
Ta có:
$\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$
$=a\cdot \dfrac{a}{b+c}+b\cdot\dfrac{b}{c+a}+c\cdot\dfrac{c}{a+b}$
$=a\cdot\left(\dfrac{a}{b+c}+1-1\right)+b\cdot\left(\dfrac{b}{c+a}+1-1\right)+c\cdot\left(\dfrac{c}{a+b}+1-1\right)$
$=a\cdot\left(\dfrac{a+b+c}{b+c}-1\right)+b\cdot\left(\dfrac{b+c+a}{c+a}-1\right)+c\cdot\left(\dfrac{c+a+b}{a+b}-1\right)$
$=a\cdot\dfrac{a+b+c}{b+c}-a+b\cdot\dfrac{a+b+c}{c+a}-b+c\cdot\dfrac{a+b+c}{a+b}-c$
$=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-\left(a+b+c\right)$
$=\left(a+b+c\right)\cdot 1-\left(a+b+c\right)$
$=0$
