Решебник по алгебре 7 класс Мерзляк ФГОС Задание 710

 

\[\boxed{\text{710\ (710).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 8x² + 16\text{xy} + 8y² =\]

\[= 8 \cdot \left( x^{2} + 2\text{xy} + y^{2} \right) =\]

\[= 8 \cdot (x + y)²\]

\[2) - 2a^{2} + 24\text{ab} - 72b^{2} =\]

\[= - 2 \cdot \left( a^{2} - 12ab + 36b^{2} \right) =\]

\[= - 2 \cdot (a - 6b)²\]

\[3) - 12b^{3} - 12b^{2} - 3b =\]

\[= - 3b\left( 4b^{2} + 4b + 1 \right) =\]

\[= - 3b(2b + 1)²\]

\[4)\ 48m³n - 72m^{2}n + 27mn =\]

\[= 3mn\left( 16m^{2} - 24m + 9 \right) =\]

\[= 3mn(4m - 3)²\]

\[\boxed{\text{710.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1){\ (a + 3)}^{2} - (a - 9)(a + 9) =\]

\[= a^{2} + 6a + 9 - a^{2} + 81 =\]

\[= 6a + 90\]

\[a = - 2,5:\]

\[- 6 \cdot 2,5 + 90 = 90 - 15 = 75.\]

\[x = - \frac{1}{3}:\]

\[9 \cdot \left( - \frac{1}{3} \right)^{2} - 30 \cdot ( - \frac{1}{3} + 55 =\]

\[= 9 \cdot \frac{1}{9} + \frac{30}{3} + 55 =\]

\[= 1 + 10 + 55 = 66.\]

\[y = \frac{1}{2}:\]

\[36 \cdot \left( \frac{1}{2} \right)^{4} + 30 = 36 \cdot \frac{1}{16} + 30 =\]

\[= 30\frac{9}{4} = 32\frac{1}{4}.\]