hreury

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

\(P=a^2x+b^2y+c^2z=\left(b+c\right)^2x+\left(c+a\right)^2y+\left(a+b\right)^2z\)\(=\left(b^2x+c^2x+c^2y+a^2y+a^2z+b^2z\right)+2\left(bcx+acy+abz\right)\)\(=a^2\left(y+z\right)+b^2\left(z+x\right)+c^2\left(x+y\right)+2\left(bcx+acy+abz\right)=0\)ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\Leftrightarrow xbc+ayc+abz=0\)

\(\Rightarrow P=-a^2x-b^2y-c^2z\)

\(\Rightarrow a^2x+b^2y+c^2z=-\left(a^2x+b^2y+c^2z\right)\Rightarrow2\left(a^2x+b^2y+c^2z\right)=0\Rightarrow P=0\)

Cộng 3 vế pt ta được:

\(x+y-z+x-y+z-x-y+z=0\Leftrightarrow x+y+z=0\)

\(\text{Đặt }\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow\hept{\begin{cases}x=ak\\y=bk\\z=ck\end{cases}}\)

Khi đó : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}=\frac{a^2}{ak}+\frac{b^2}{bk}+\frac{c^2}{ck}=\frac{a}{k}+\frac{b}{k}+\frac{c}{k}=\frac{a+b+c}{k}\left(1\right);\)

\(\frac{\left(a+b+c\right)^2}{x+y+z}=\frac{\left(a+b+c\right)^2}{ak+bk+ck}=\frac{\left(a+b+c\right)^2}{k\left(a+b+c\right)}=\frac{a+b+c}{k}\left(2\right)\)

Từ (1) và (2) => \(\frac{a^2}{x}+\frac{b^2}{y}=\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\left(\text{đpcm}\right)\)

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