Решебник по алгебре 7 класс Мерзляк Задание 416

\[\boxed{\text{416\ (416).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (a - 2) \cdot \ (*\ + \ 6) =\]

\[= a^{2} + \ * - *\]

\[(a - 2) \cdot \ \left( *_{1}\ + \ 6 \right) =\]

\[= a^{2} + \ *_{2} - *_{3}\]

\[a \cdot *_{1} - 2 \cdot *_{1} + 6a - 12 =\]

\[= a^{2} + *_{2} - *_{3}\]

\[a \cdot *_{1} = a^{2}\]

\[*_{1} = \frac{a^{2}}{a} = a.\]

\[a^{2} - 2a + 6a - 12 =\]

\[= a^{2} + *_{2} - *_{3}\]

\[a^{2} + 4a - 12 = a^{2} + *_{2} - *_{3}\]

\[*_{1} = a;\ \ \ \ \ *_{2} = 4a;\ \ \ *_{3} = 12.\]

\[2)\ (2a + 7) \cdot (a - *) =\]

\[= \ *\ + \ * - 14\]

\[(2a + 7) \cdot \left( a - *_{1} \right) =\]

\[= \ *_{2}\ + \ *_{3} - 14\]

\[2a^{2} + 7a - 2a \cdot *_{1} - 7 \cdot *_{1} =\]

\[= *_{2} + *_{3} - 14\ \]

\[- 7*_{1} = - 14\]

\[*_{1} = - 14\ :( - 7)\]

\[*_{1} = 2.\]

\[2a^{2} + 7a - 4a - 14 =\]

\[= *_{2} + *_{3} - 14\]

\[2a^{2} + 3a - 14 = *_{2} + *_{3} - 14\ \]

\[*_{1} = 2;\ \ \ *_{2} = 2a^{2};\ \ \ *_{3} = 3a.\]