<=> (z² + 4z+4 )+(2x²-4x +2 )+(3x² +6xy +3y²)=0 
<=> (z+2)² +2(x-2)² +3(x+y)²=0 
ba hằng đẳng thức=> ta được: z+2 =0 và x-2=0 và x+y= 0 
=> z=-2, x=2 , y= -2 

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

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

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

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

Vậy x=-1 y=1

a)  \(x^2+2y^2+2xy-2y+1=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)

b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)

\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)

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

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

            \(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)

a) \(x^2+2y^2+2xy-2y+1=0\)

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

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

Vậy x=-1 ; y=1

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

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

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

Ta có : x² + 4y² - 2x + 4y + 2 = 0

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

<=> (x - 1)² + (2x + 1)² = 0

Mà : \(\left(x-1\right)^2\ge0\forall x\)

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

Nên \(\orbr{\begin{cases}x-1=0\\2x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)

Bài 1:

\(a,=\left(156-56\right)^2=100^2=10000\\ b,=\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)\)

Bài 2:

\(a,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ b,=x^2+2x-6x-12=\left(x+2\right)\left(x-6\right)\\ c,=3\left(x^2-2xy-16+y^2\right)=3\left[\left(x-y\right)^2-16\right]\\ =3\left(x-y-4\right)\left(x-y+4\right)\)