Question

Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\). Chứng minh rằng: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

Answer 1

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\) (1)

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+ayz+bxz}{abc}\right)=1\)

Kết hợp với (1) ta có đpcm