Violympic toán 8

Lucy Heartfilia

Cho a,b,c thỏa mãn

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\)\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)

Chứng minh : a=b=c

Nguyễn Việt Lâm
2 tháng 1 2019 lúc 20:19

\(\Leftrightarrow a^2\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+b^2\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)+c^2\left(\dfrac{1}{c+a}-\dfrac{1}{a+b}\right)=0\)

\(\Leftrightarrow\dfrac{a^2\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^2\left(a-b\right)}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^2\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}=0\)

\(\Leftrightarrow a^2\left(c-a\right)\left(c+a\right)+b^2\left(a-b\right)\left(a+b\right)+c^2\left(b-c\right)\left(b+c\right)=0\)

\(\Leftrightarrow a^2\left(c^2-a^2\right)+b^2\left(a^2-b^2\right)+c^2\left(b^2-c^2\right)=0\)

\(\Leftrightarrow a^2c^2+a^2b^2+b^2c^2-a^4-b^4-c^4=0\)

\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2b^2-2a^2c^2-2b^2c^2=0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a^2-c^2\right)^2+\left(b^2-c^2\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b^2=0\\a^2-c^2=0\\b^2-c^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a+b\right)=0\\\left(a-c\right)\left(a+c\right)=0\\\left(b-c\right)\left(b+c\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\) (do \(\left(a+b\right)\left(a+c\right)\left(b+c\right)\ne0\) \(\Rightarrow\left\{{}\begin{matrix}a+b\ne0\\a+c\ne0\\b+c\ne0\end{matrix}\right.\))

\(\Rightarrow a=b=c\)

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