\[\boxed{\text{64\ (64).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 6x - 2 = x^{2} - 6x + 9 -\]
\[- 11 = \left( x^{2} - 6x + 9 \right) - 11 =\]
\[= (x - 3)^{2} - 11\]
\[\textbf{б)}\ x^{2} + 5x + 20 = x^{2} + 5x +\]
\[+ {2,5}^{2} - {2,5}^{2} + 20 =\]
\[= (x + 2,5)^{2} + 13,75\]
\[\textbf{в)}\ 2x^{2} - 4x + 10 =\]
\[= 2 \cdot \left( x^{2} - 2x + 5 \right) =\]
\[= 2 \cdot \left( x^{2} - 2x + 1 + 4 \right) =\]
\[= 2 \cdot \left( x^{2} - 2x + 1 \right) + 2 \cdot 4 =\]
\[= 2 \cdot (x - 1)^{2} + 8\]
\[\textbf{г)}\ \frac{1}{2}x^{2} + x - 6 =\]
\[= \frac{1}{2} \cdot \left( x^{2} + 2x - 12 \right) =\]
\[= \frac{1}{2} \cdot \left( x^{2} + 2x + 1 - 13 \right) =\]
\[= \frac{1}{2} \cdot \left( x^{2} + 2x + 1 \right) - \frac{1}{2} \cdot 13 =\]
\[= \frac{1}{2} \cdot (x + 1)^{2} - 6,5.\]