Question

tìm x,y,z biết

x^2 + 4x + y^2 -8y + 4z^2 +4z +21 = 0

Answer 1

x² + 4x + y²  - 8y + 4z² + 4z + 21 = 0

<=> (x² + 4x + 4) + (y² - 8y + 16) + (4z² + 4z + 1) = 0

<=> (x + 2)² + (y - 4)² + (2z + 1)² = 0

<=> \(\hept{\begin{cases}x+2=0\\y-4=0\\2z+1=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-2\\y=4\\z=-\frac{1}{2}\end{cases}}\)

Answer 2

x² + 4x + y² - 8y + 4z² + 4z + 21 = 0

⇔ ( x² + 4x + 4 ) + ( y² - 8y + 16 ) + ( 4z² + 4z + 1 ) = 0

⇔ ( x + 2 )² + ( y - 4 )² + ( 2z + 1 )² = 0

⇔ \(\hept{\begin{cases}x+2=0\\y-4=0\\2z+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=4\\z=-\frac{1}{2}\end{cases}}\)