Giải thích các bước giải:
1,
\[\begin{array}{l}
1.{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\\
2.{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\\
3.{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\\
4.{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\\
5.{\left( {a - b} \right)^3} = {a^3} - 3{a^2}b + 3a{b^2} - {b^3}\\
6.{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
7.\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
2,\\
a,{x^2} + 2x + 1 = {\left( {x + 1} \right)^2}\\
b,9{x^2} + {y^2} + 6xy = {\left( {3x} \right)^2} + 2.3x.y + {y^2} = {\left( {3x + y} \right)^2}\\
c,25{a^2} + 4{b^2} - 20ab = {\left( {5a} \right)^2} - 2.5a.2b + {\left( {2b} \right)^2} = {\left( {5a - 2b} \right)^2}\\
d,{x^2} - x + \frac{1}{4} = {\left( {x - \frac{1}{2}} \right)^2}
\end{array}\]
