Ta có :
+) \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow\)\(\frac{ayz+bxz+cxy}{xyz}=0\) \(\left(1\right)\)
Lại có :
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=0\) \(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
Đặt \(A=\frac{x}{a},B=\frac{y}{b};C=\frac{z}{c}\)
Theo đề bài : \(\left\{{}\begin{matrix}\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=0\\A+B+C=1\end{matrix}\right.\)
Lại có : \(\left(A+B+C\right)^2=A^2+B^2+C^2+2.\left(AB+BC+CA\right)\)
\(=A^2+B^2+C^2+2ABC\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)\)
\(=A^2+B^2+C^2=1\)
Hay \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1_{đpcm}\)
