x.(x+2y) = -3

x, y nguyên nên ta có bảng sau

Kết luận nhé

Viết lại bài này đi

......................?

mik ko biết

mong bn thông cảm 

nha ................

a) x²+2y²+2xy-2y+1=0

\(\Leftrightarrow\)(x²+2xy+y²)+(y²-2y+1)=0

\(\Leftrightarrow\)(x+y)²+(y-1)²=0

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

Vậy x=-1, y=1

a/ \(x^2+2y^2+2xy-2y+1=0\)

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

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

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

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

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

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

b/ \(x^2+2y^2+2xy-2x+2=0\)

<=> \(\left(x^2+2xy+y^2\right)+\left(2y-2x+2\right)=0\)

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

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

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

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

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

Trừ (1) và (2)

=> \(2y=-1\)

<=> \(y=-\frac{1}{2}\)

<=> \(x=\frac{1}{2}\)(vì \(x+y=0\)<=> \(x=-y\))

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) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)

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

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

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

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

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

(x-y)²=0

x-y=0

x=y

voi moi so nguyen x,y

x² - 2xy + y² = 0

x² - xy - xy + y² = 0

x(x - y) - y(x - y) = 0

(x - y)(x - y) = 0

(x - y)² = 0

x - y = 0

=> x = y

Vậy x = y thì x² - 2xy + y² = 0