Giúp mình với !!!!!!!!!!!!!!

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {x + y – 2} \right)^2} + 2.\left( {x + y – 2} \right).\left( {x – y + 2} \right) + {\left( {x – y + 2} \right)^2}\\
 = {\left[ {\left( {x + y – 2} \right) + \left( {x – y + 2} \right)} \right]^2}\\
 = {\left( {2x} \right)^2}\\
 = 4{x^2}\\
b,\\
\left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right) + \left( {x – y} \right)\left( {{x^2} + xy + {y^2}} \right)\\
 = \left( {{x^3} + {y^3}} \right) + \left( {{x^3} – {y^3}} \right)\\
 = 2{x^3}\\
2,\\
a,\\
2{x^2} + 4xy – 4x + 4 + 4{y^2} = 0\\
 \Leftrightarrow \left( {{x^2} + 4xy + 4{y^2}} \right) + \left( {{x^2} – 4x + 4} \right) = 0\\
 \Leftrightarrow {\left( {x + 2y} \right)^2} + {\left( {x – 2} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2y} \right)^2} = 0\\
{\left( {x – 2} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y =  – 1
\end{array} \right.\\
b,\\
{x^2} + 2{y^2} – 2y + 1 – 2xy = 0\\
 \Leftrightarrow \left( {{x^2} – 2xy + {y^2}} \right) + \left( {{y^2} – 2y + 1} \right) = 0\\
 \Leftrightarrow {\left( {x – y} \right)^2} + {\left( {y – 1} \right)^2} = 0\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – y} \right)^2} = 0\\
{\left( {y – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = y = 1
\end{array}\)