\[\boxed{\text{799\ (799).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ (x + y)^{2} = x^{2} + 2xy + y^{2}\]
\[\textbf{б)}\ (p - q)^{2} = p^{2} - 2pq + q^{2}\]
\[\textbf{в)}\ (b + 3)^{2} = b^{2} + 2 \cdot 3b + 3^{2} =\]
\[= b^{2} + 6b + 9\]
\[\textbf{г)}\ (10 - c)^{2} =\]
\[= 10^{2} - 2 \cdot 10c + c^{2} =\]
\[= 100 - 20c + c^{2}\]
\[\textbf{д)}\ (y - 9)^{2} = y^{2} - 2 \cdot 9y + 9^{2} =\]
\[= y^{2} - 18y + 81\]
\[\textbf{е)}\ (9 - y)^{2} = 9^{2} - 2 \cdot 9y + y^{2} =\]
\[= 81 - 18y + y^{2}\]
\[\textbf{ж)}\ (a + 12)^{2} =\]
\[= a^{2} + 2 \cdot 12a + 12^{2} =\]
\[= a^{2} + 24a + 144\]
\[\textbf{з)}\ (15 - x)^{2} =\]
\[= 15^{2} - 2 \cdot 15x + x^{2} =\]
\[= 225 - 30x + x^{2}\]
\[\textbf{и)}\ (b - 0,5)^{2} =\]
\[= b^{2} - 2 \cdot 0,5b + {0,5}^{2} =\]
\[= b^{2} - b + 0,25\]
\[к)\ (0,3 - m)^{2} =\]
\[= {0,3}^{2} - 2 \cdot 0,3m + m^{2} =\]
\[= 0,09 - 0,6m + m^{2}\ \]