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Bài này thiếu điều kiện $x,y,z \not= \, 0$ 
Ta có: $\dfrac{(ax+by+cz)^2}{x^2+y^2+z^2}=\dfrac{2304}{64}=36=a^2+b^2+c^2$
$\rightarrow a^2+b^2+c^2=\dfrac{(ax+by+cz)^2}{x^2+y^2+z^2}$
$\leftrightarrow (a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2$
$\leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz$
$\leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2axcz-2bycz=0$
$\leftrightarrow (ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$
$\leftrightarrow \left\{\begin{matrix}ay-bx=0\\ az-cx=0\\ bz-cy=0\end{matrix}\right. \leftrightarrow \left\{\begin{matrix}ay=bx\\ az=cx\\ bz=cy\end{matrix}\right.$
$\leftrightarrow \left\{\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\ \dfrac{a}{x}= \dfrac{c}{z} \\ \dfrac{b}{y}= \dfrac{c}{z} \end{matrix}\right. \leftrightarrow \dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}$
$\rightarrow \dfrac{(a+b+c)^2}{(x+y+z)^2}=\dfrac{a^2+b^2+c^2}{x^2+y^2+z^2}=\dfrac{36}{64}$
$\rightarrow \dfrac{a+b+c}{x+y+z}=\dfrac{6}{8}$
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