cho a,b,c thảo mãn a/(b+c) + b/(c+a) +c/(a+b)=1 tính a^2/(b+c) + b^2/c+a + c^2 (a+b)

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1. 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:

   $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$

   $=a\cdot \dfrac{a}{b+c}+b\cdot\dfrac{b}{c+a}+c\cdot\dfrac{c}{a+b}$

   $=a\cdot\left(\dfrac{a}{b+c}+1-1\right)+b\cdot\left(\dfrac{b}{c+a}+1-1\right)+c\cdot\left(\dfrac{c}{a+b}+1-1\right)$

   $=a\cdot\left(\dfrac{a+b+c}{b+c}-1\right)+b\cdot\left(\dfrac{b+c+a}{c+a}-1\right)+c\cdot\left(\dfrac{c+a+b}{a+b}-1\right)$

   $=a\cdot\dfrac{a+b+c}{b+c}-a+b\cdot\dfrac{a+b+c}{c+a}-b+c\cdot\dfrac{a+b+c}{a+b}-c$

   $=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-\left(a+b+c\right)$

   $=\left(a+b+c\right)\cdot 1-\left(a+b+c\right)$
   $=0$

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