Question

Tìm x , y :

a) x^2 + y^2 + 10x + 6y + 34 = 0

b) 25x^2 + 4y^2 + 10x + 4y + 2 = 0

Answer 1

x² + y² + 10x + 6y + 34 = 0

=> (x² + 10x + 25) + (y² + 6y + 9) = 0

=> (x + 5)² + (y + 3)² = 0

=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

Vậy x = - 5 ; y = -3

b) 25x² + 4y² + 10x + 4y + 2 = 0

=> (25x² + 10x + 1) + (4y² + 4y + 1) = 0

=> (5x + 1)² + (2y + 1)² = 0

=> \(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-0,2\\y=-0,5\end{cases}}\)

Vậy x = -0,2 ; y = -0,5

Answer 2

a) 

\(x^2+10x+25+y^2+6y+9=0\)    

\(\left(x+5\right)^2+\left(y+3\right)^2=0\)  ( 1 ) 

Ta có : 

\(\left(x+5\right)^2\ge0\forall x\) 

\(\left(y+3\right)^2\ge0\forall y\) 

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\)         

\(\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)   

b) 

\(25x^2+10x+1+4y^2+4y+1=0\)     

\(\left(5x+1\right)^2+\left(2y+1\right)^2=0\) ( 1 ) 

Ta có : 

\(\left(5x+1\right)^2\ge0\forall x\)      

\(\left(2y+1\right)^2\ge0\forall y\)  

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(5x+1\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\)    

\(\hept{\begin{cases}x=\frac{-1}{5}\\y=\frac{-1}{2}\end{cases}}\)

Answer 3

x² + y² + 10x + 6y + 34 = 0

<=> ( x² + 10x + 25 ) + ( y² + 6y + 9 ) = 0

<=> ( x + 5 )² + ( y + 3 )² = 0

<=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

25x² + 4y² + 10x + 4y + 2 = 0

<=> ( 25x² + 10x + 1 ) + ( 4y² + 4y + 1 ) = 0

<=> ( 5x + 1 )² + ( 2y + 1 )² = 0

<=> \(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{5}\\y=-\frac{1}{2}\end{cases}}\)