\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\) chú thích: (2)=(1)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1-2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)\) (2)
tìm \(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\)ta có
\(\Leftrightarrow\dfrac{xyc+yza+xzb}{abc}\) (1)
từ \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
thay vào (1)
\(\Rightarrow\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}=0\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2011=2012\)
