a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

\(3x^2+3y^2+4xy+2x-2y+2=0\)

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

\(\Leftrightarrow2\left(x^2+y^2+2xy\right)+\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

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

\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-y\\x=-1\\y=1\end{cases}}\)

Vậy \(x=-1\)và \(y=1\)

Sửa đề: \(3x^2+3y^2+4xy+2x-2y+2=0\)

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

=>\(2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
=>\(\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

\(3x^2y-12x^3y^2=3x^2y\left(1-4xy\right)\)

\(2x^2\left(x-y\right)-4xy\left(x-y\right)\)

\(=2x\left(x-y\right)\left(x-2y\right)\)

a) Ta có: A = x² + y² - xy - 2x - 2y + 9

2A = 2x² + 2y² - 2xy - 4x - 4y + 18

2A = (x² + y² - 2xy) + (x² - 4x + 4) + (x² - 4y + 4) + 10

2A = (x - y)² + (x - 2)² + (y - 2)² + 10 \(\ge\)10 \(\forall\)x

=>A \(\ge\)5 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\x-2=0\\y-2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=y\\x=2\\y=2\end{cases}}\) <=> x = y = 2

Vậy MinA = 5 <=> x = y = 2

b) Ta có: 3x² + 3y² + 4xy + 2x - 2y + 2 = 0

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

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

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

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

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