<=>4x²+8xy+4y² +x²-2x+1+y²+2y+1=0

<=>(2x+2y)²+(x-1)²+(y+1)²=0

<=>(2x+2y)²=0 và (x-1)²=0 và (y+1)²=0

*(x-1)²=0

<=> x-1=0

<=>x=1

*(y+1)²

<=> y+1=0

<=> y=-1

Vậy x=1;y= -1

Ta có: x^2+2y^2-2xy+2x+2-4y=0

=> x^2 -2xy+y^2+ 2x-2y+1+y^2-2y+1=0

=> (x-y)^2+ 2(x-y)+1 + (y-1)^2=0

=> (x-y+1)^2+(y-1)^2=0

mà (x-y+1)^2> hoặc=0 với mọi x;y

(y-1)^2> hoặc=0 với mọi x;y

nên x-y+1=0;y-1=0

=> y=1; x=0

=> x²-2x+1+y²+2y+1+4x²+8xy+4y²=0

=>(x-1)²+(y+1)²+(2x+2y)²=0

=>x-1=0 va y+1=0 va 2x+2y=0

=>x=1 va y=-1

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

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

\(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Vì      \(\left(x+y\right)^2\ge0;\left(x-1\right)^2\ge0;\left(y+1\right)^2\ge0\)

Để    \(4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow x+y=0\)

\(\Leftrightarrow y+1=0\Rightarrow y=-1\)

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

Vậy    \(x=1; y=-1\)

a)\(x^2+5y^2-2xy+4y+1=0\)

\(x^2+2xy+y^2+4y^2+4y+1=0\)

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

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

      Từ (1) ta đc: x = 1/2

b)\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

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

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

CÂU B Sao bạn làm được vậy

Bài làm:

a) \(x^2+5y^2-2xy+4y+1=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)=0\)

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

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(2y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)

Vậy \(x=y=-\frac{1}{2}\)

b) \(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(x+1\right)^2=0\)

Vì \(\hept{\begin{cases}4\left(x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}4\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)

<=>4x²+8xy+4y² +x²-2x+1+y²+2y+1=0

<=>(2x+2y)²+(x-1)²+(y+1)²=0

<=>(2x+2y)²=0 và (x-1)²=0 và (y+1)²=0

*(x-1)²=0

<=> x-1=0

<=>x=1

*(y+1)²

<=> y+1=0

<=> y=-1

Vậy x=1;y= -1

 5x^2+5y^2+8xy-2x+2y+2 = 0 
<=>4x^2 + 8xy + 4y^2 + x^2 - 2x + 1 + y^2 + 2y + 1 = 0 
<=> 4(x + y)^2 + (x - 1)^2 + (y + 1)^2 = 0 (1) 
mà 4(x + y)^2 >= 0;(x - 1)^2 >=0; (y + 1)^2 >= 0 
=> Để (1) có nghiệm thì đồng thời x + y = 0; x - 1 = 0; y + 1 = 0 
<=> x = 1, y = -1.