Question

Tìm x,y,z thỏa mãn phương trình sau:

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

Answer 1

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

( 9x² -18x + 9) +( y² - 6y + 9) +2(z²+2z +1) = 0

( 3x-3)² + ( y-3)² + 2( z+1)² = 0

vì ( 3x-3)^2 , (y-3)^2 , 2( z+1)^2 >0 \(\Rightarrow\left(3x-3\right)^2=\left(y-3\right)^2=2\left(z+1\right)^2\))^2

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