Câu hỏi của Nguyễn Thu Hằng

Question

Cho $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1$. Tính $\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}$

Đinh Đức Hùng · Accepted Answer

Từ $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1$

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

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

$\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0$Câu trả lời: Từ $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1$

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

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

$\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0$

hattori heiji · Answer

$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1$

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

$\Leftrightarrow\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{c+a}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c$$\Leftrightarrow\dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c$

$\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c-b-a-c=0$

$\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$Câu trả lời: $\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1$

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

$\Leftrightarrow\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{c+a}+\dfrac{c^2+c\left(a+b\right)}{a+b}=a+b+c$$\Leftrightarrow\dfrac{a^2}{b+c}+a+\dfrac{b^2}{c+a}+b+\dfrac{c^2}{a+b}+c=a+b+c$

$\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c-b-a-c=0$

$\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0$

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