![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
<=>(x2+y2+z2+2xy+2yz+2xz)+(x2+2x+1)+(y2+4y+4)=0
<=>(x+y+z)2+(x+1)2+(y+2)2=0
Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(y+2\right)^2\ge0}\)
=>\(\hept{\begin{cases}x+y+z=0\\x+1=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}z=3\\x=-1\\y=-2\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 + y2 - 2x +4y + 5 = 0
⇔ ( x2 - 2x +1) + ( y2 + 4y + 4 ) = 0
⇔ ( x - 1)2 + ( y + 2)2 = 0
Do : ( x-1)2 ≥ 0 , ( y + 2 )2 ≥ 0
⇒ ( x - 1 )2 = 0 và (y+2)2 = 0
⇒ x = 1 và y = -2
\(x^2+y^2-2x+4y+5=0\)
=>\(x^2-2x+y^2+4y+5=0\)
=>\(x^2-2x1+1-1+y^2+2y2+2^2+1=0\)
=>\(\left(x-1\right)^2\left(y+2\right)^2=0\)
Vì\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^2-2x+1) + (y^2+4y+4) = 0
(x-1)^2 + (y+2)^2 = 0
Suy ra x-1 = 0 và y +2 = 0
x = 1 và y = -2
Ta có \(x^2-2x+y^2+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x2 + y2 + 2x - 4y + 5 = 0
<=> ( x2 + 2x +1 ) + ( y2 - 4y + 4 ) = 0
<=> ( x + 1 )2 + ( y - 2 ) 2 = 0
<=> \(\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x+1=0\\y-2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-1\\y=2\end{cases}}\)
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
<=> \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\2y+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{2}\\2y=-1\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-1}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2+y2-2x-4y+5=0
<=>(x2-2x+1)+(y2-4y+4)=0
<=>(x-1)2+(y-2)2=0
tổng 2 số ko âm=0 <=>chúng đều=0
<=>x=1;y=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x^2+y^2-2x+4y+5=0\)
<=> \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
<=> \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
Vì \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\)
=> \(\left[\begin{array}{nghiempt}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x-1=0\\y+2=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=1\\y=-2\end{array}\right.\)
Vậy x=1 ; y=-2
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
Ta thấy: \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\forall x;y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Do đó: \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}\). Vậy ...