Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
{\left( {x + y} \right)^2} - 2.\left( {{x^2} - {y^2}} \right) + {\left( {x - y} \right)^2}\\
= {\left( {x + y} \right)^2} - 2.\left( {x + y} \right)\left( {x - y} \right) + {\left( {x - y} \right)^2}\\
= {\left[ {\left( {x + y} \right) - \left( {x - y} \right)} \right]^2}\\
= {\left( {2y} \right)^2}\\
= 4{y^2}\\
*)\\
M = x.\left( {{x^2} - 3} \right) - {\left( {x - 1} \right)^3} + 2\\
= \left( {{x^3} - 3x} \right) - \left( {{x^3} - 3{x^2} + 3x - 1} \right) + 2\\
= {x^3} - 3x - {x^3} + 3{x^2} - 3x + 1 + 2\\
= 3{x^2} - 6x + 3\\
= 3.\left( {{x^2} - 2x + 1} \right)\\
= 3.{\left( {x - 1} \right)^2}\\
N = {\left( {xy - 1} \right)^2} - 2.\left( {xy - 1} \right)\left( {xy + 2} \right) + {\left( {xy + 2} \right)^2}\\
= {\left[ {\left( {xy - 1} \right) - \left( {xy + 2} \right)} \right]^2}\\
= {\left( { - 3} \right)^2}\\
= 9
\end{array}\)
