a) x2 + y2 - 6x + 2y + 10 = 0
<=> ( x2 - 6x + 9 ) + ( y2 + 2y + 1 ) = 0
<=> ( x - 3 )2 + ( y + 1 )2 = 0
<=> \(\hept{\begin{cases}x-3=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)
b) 4x2 + y2 - 20x - 2y + 26 = 0
<=> ( 4x2 - 20x + 25 ) + ( y2 - 2y + 1 ) = 0
<=> ( 2x - 5 )2 + ( y - 1 )2 = 0
<=> \(\hept{\begin{cases}2x-5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=1\end{cases}}\)
a) x2 + y2 - 6x + 2y + 10 = 0
=> (x2 - 6x + 9) + (y2 + 2y + 1) = 0
=> (x - 3)2 + (y + 1)2 = 0 (1)
Vì \(\hept{\begin{cases}\left(x-3\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-3\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Đẳng thức (1) xảy ra <=> \(\hept{\begin{cases}x-3=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}\)
Vậy x = 3 ; y = -1
b) 4x2 + y2 + 20x - 2y + 26 = 0
=> (4x2 - 20x + 25) + (y2 - 2y + 1) = 0
=> (2x - 5)2 + (y - 1)2 = 0 (1)
Vì \(\hept{\begin{cases}\left(2x-5\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(2x-5\right)^2+\left(y-1\right)^2\ge0\forall x;y\)
Đẳng thức (1) "=" xảy ra <=> \(\hept{\begin{cases}2x-5=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2,5\\y=1\end{cases}}\)
Vậy x = 2,5 ; y = 1
\(\text{a)}\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+1\right)^2=0\)
\(\text{Vì }\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}\text{nên }\left(x-3\right)^2}+\left(y+1\right)^2\ge0\)
\(\text{Dấu = xảy ra }\Leftrightarrow\hept{\begin{cases}x-3=0\\y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=-1\end{cases}}\)
\(\text{b)}\Leftrightarrow\left(4x^2-20x+25\right)+\left(y^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2+\left(y-1\right)^2=0\)
\(\text{Tương tự phần a ,}\Rightarrow\hept{\begin{cases}2x-5=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=1\end{cases}}\)
