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: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

Giúp hộ!!!

Answer 1

Ta có :

+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

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

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

\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)