Решебник по алгебре 7 класс Мерзляк ФГОС Задание 417

 

\[\boxed{\text{417\ (417).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (x + 3) \cdot (* + \ 5) =\]

\[= 3x^{2} + \ * + \ *\]

\[(x + 3) \cdot \left( *_{1} + \ 5 \right) =\]

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

\[x \cdot *_{1} + x \cdot 5 + 3 \cdot *_{1} + 3 \cdot 5 =\]

\[= 3x^{2} + \ * + \ *\]

\[x \cdot *_{1} + 5x + 3 \cdot *_{1} + 15 =\]

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

\[x \cdot *_{1} = 3x^{2}\]

\[*_{1} = 3x.\]

\[5x + 3 \cdot *_{1} = 5x + 3 \cdot 3x =\]

\[= 5x + 9x = 14x = *_{2}\]

\[(x + 3)(3x + 5) =\]

\[= 3x^{2} + 14x + 15;\ \ *_{3} = 15.\]

\[*_{1} = 3x;\ \ \ *_{2} = 14x;\ \ \ *_{3} = 15.\]

\[2)\ (x - 4) \cdot (x + \ *) =\]

\[= \ *\ + \ *\ + 24\]

\[(x - 4) \cdot \left( x + \ *_{1} \right) =\]

\[= \ *_{2}\ + \ *_{3}\ + 24\]

\[x \cdot x + x \cdot *_{1} - 4 \cdot x - 4 \cdot *_{1} =\]

\[= *_{2}\ + \ *_{3}\ + 24\]

\[x^{2} + x \cdot *_{1} - 4x - 4 \cdot *_{1} =\]

\[= *_{2}\ + \ *_{3}\ + 24\]

\[- 4 \cdot *_{1} = 24\]

\[*_{1} = - 6.\]

\[x \cdot *_{1} - 4x = x \cdot ( - 6) - 4x =\]

\[= - 6x - 4x = - 10x = *_{3}.\]

\[x^{2} = *_{3}.\]

\[(x - 4)(x - 6) =\]

\[= x^{2} - 10x + 24\]

\[*_{1} = 6;\ \ \ *_{2} = x^{2};\ \ \ \ *_{3} = - 10x.\]

\[\boxed{\text{417.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \left( 5a^{3} - 20a^{2} \right) - \left( 4a^{3} - 18a^{2} \right) =\]

\[= 5a^{3} - 20a^{2} - 4a^{3} + 18a^{2} =\]

\[= a^{3} - 2a^{2}.\]

\[При\ a = - 3:\]

\[a^{3} - 2a^{2} = ( - 3)^{3} - 2 \cdot ( - 3)^{2} =\]

\[= - 27 - 2 \cdot 9 = - 27 - 18 =\]

\[= - 45\]

\[Ответ:\ - 45.\]

\[При\ b = - 1,5;\ \ c = 4:\]

\[- 4 \cdot ( - 1,5) \cdot 4 = 24.\]

\[Ответ:24.\ \]