mk sửa lại đoạn sau:

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

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

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

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

Ta có:

3x²+4y²+5z²+4xy-4yz-6xz-2x-4y-2z+3=0

<=> 4y²+4yx-4yz-4y+x²-2xz+z²-2x+2z+1+4z²-4xz-4z+x²+2x+1+x²-2x+1=0

<=>4y²+4y(x-z-1)+(x-z)²-2(x-z)+1+4z²-4z(x+1)+(x+1)²+(x-1)²=0

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

<=>(2y+x-z-1)²+(2z-x-1)²+(x-1)²=0

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

Vay....

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

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

cái đây mà Toán 6 ak ????

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