Question

Giải câu b ạ

Answer 1

\(a)9x^2+y^2+2z^2-18x+4z-6y+20=0 \\<=>9(x-1)^2+(y-3)^2+2(z+1)^2=0 \\->x-1=0; y-3=0; z+1=0 \\-> x=1; y=3; z=-1 \\b)\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0 \\<=> ayz+bxz+cxy=0 \\\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1 \)

\(\\<=>\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac})=1 \\<=>\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{ayz+bxz+cxy}{abc}=1 \\<=>\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1 (đpcm)\)