Решебник по алгебре 7 класс Мерзляк ФГОС Задание 708

 

\[\boxed{\text{708\ (708).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 12b² - 12c^{2} =\]

\[= 12 \cdot \left( b^{2} - c^{2} \right) =\]

\[= 12 \cdot (b - c)(b + c)\]

\[2)\ 2a²c - 2b^{2}c =\]

\[= 2c\left( a^{2} - b^{2} \right) =\]

\[= 2c(a - b)(a + b)\]

\[3)\ 5a² - 20 = 5 \cdot \left( a^{2} - 4 \right) =\]

\[= 5 \cdot (a - 2)(a + 2)\]

\[4)\ 3mn² - 48m =\]

\[= 3m\left( n^{2} - 16 \right) =\]

\[= 3m(n - 4)(n + 4)\]

\[5)\ 7y³ - 7y = 7y\left( y^{2} - 1 \right) =\]

\[= 7y(y - 1)(y + 1)\]

\[6)\ a³ - a^{5} = a^{3}\left( 1 - a^{2} \right) =\]

\[= a³(1 - a)(1 + a)\ \]

\[\boxed{\text{708.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 6 \cdot (1 - 2c)^{2} =\]

\[= 6 \cdot \left( 1 - 4c + 4c^{2} \right) =\]

\[= 6 - 24c + 24c^{2}\]

\[2) - 12 \cdot \left( x + \frac{1}{3}y \right)^{2} =\]

\[= - 12 \cdot \left( x^{2} + \frac{2}{3}xy + \frac{1}{9}y^{2} \right) =\]

\[= - 12x^{2} - 8xy - \frac{4}{3}y^{2}\]

\[3)\ a(a - 6b)^{2} =\]

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

\[= a^{3} - 12a^{2}b + 36ab^{2}\]

\[4)\ 5b\left( b^{2} + 7b \right)^{2} =\]

\[= 5b\left( b^{4} + 14b^{3} + 49b^{2} \right) =\]

\[= 5b^{5} + 70b^{4} + 245b^{3}\]

\[5)\ (a + 3)(a - 4)^{2} =\]

\[= (a + 3)\left( a^{2} - 8a + 16 \right) =\]

\[= \ a^{3} - 5a^{2} - 8a + 48\]

\[6)\ (2x + 4)^{2}(x - 8) =\]

\[= \left( 4x^{2} + 16x + 16 \right)(x - 8) =\]

\[= 4x^{3} - 16x^{2} - 112x - 128\]

\[7)\ (a - 5)^{2}(a + 5)^{2} =\]

\[= \left( (a - 5)(a + 5) \right)^{2} =\]

\[= \left( a^{2} - 25 \right)^{2} =\]

\[= a^{4} - 50a^{2} + 625\]

\[8)\ (3x + 4y)^{2}(3x - 4y)^{2} =\]

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

\[= \left( 9x^{2} - 16y^{2} \right)^{2} =\]

\[= 81x^{4} - 288x^{2}y^{2} + 256y^{4}\]