Question

[๖ACE✪Şнαdσωッ]

a. tìm x, y, z thỏa mãn phương trình sau: 

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

b. cho \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) . cmr \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)

giúp mình vs

Answer 1

[Trí Tiên]

a) Ta có :

\(9x^2+y^2+2z^2-18x+4z-6y+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+2\left(z^2+2z+1\right)=0\)

\(\Leftrightarrow\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Ta thấy : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2\ge0\forall x,y,z\)

Do đó : \(\left(3x-3\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(3x-3\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\\z=-1\end{cases}}\) ( thỏa mãn )

Vậy : \(\left(x,y,z\right)=\left(1,3,-1\right)\)

Answer 2

[๖ACE✪Şнαdσωッ]

Câu (b) nữa Vinh ơi