Question

[kudo shinichi]

Tìm giá trị của x,y biết : 

a, x² + y² + 2x - 4y +5 = 0

b, x² + 4y² - x + 4y + 5/4 = 0

Answer 1

[Sofia Nàng]

       a) x² + y² + 2x - 4y + 5 = 0

 <=> ( x² + 2x +1 ) + ( y² - 4y + 4 ) = 0

 <=> ( x + 1 )² + ( y - 2 ) ² = 0

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

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

 <=> \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

Answer 2

[Sofia Nàng]

   b) x² + 4y² - x + 4y + \(\frac{5}{4}\)=0

<=> ( x² - 2x + \(\frac{1}{4}\)) + ( 4y² + 4y + 1 ) = 0

<=> ( x - \(\frac{1}{2}\))²  + ( 2y + 1 )² = 0

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

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

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

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