Решебник по алгебре 7 класс Мерзляк ФГОС Задание 707

 

\[\boxed{\text{707\ (707).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 2a² - 2b^{2} = 2 \cdot \left( a^{2} - b^{2} \right) =\]

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

\[2)\ \text{cx}² - cy^{2} = c\left( x^{2} - y^{2} \right) =\]

\[= c(x - y)(x + y)\]

\[3)\ 3x² - 3 = 3 \cdot \left( x^{2} - 1 \right) =\]

\[= 3 \cdot (x - 1)(x + 1)\]

\[4)\ 3ab² - 27a = 3a\left( b^{2} - 9 \right) =\]

\[= 3a(b - 3)(b + 3)\]

\[5)\ 2y³ - 18y = 2y\left( y^{2} - 9 \right) =\]

\[= 2y(y - 3)(y + 3)\]

\[6)\ x³ - 4x = x\left( x^{2} - 4 \right) =\]

\[= x(x - 2)(x + 2)\]

\[7)\ x^{4} - x^{2} = x^{2}\left( x^{2} - 1 \right) =\]

\[= x²(x - 1)(x + 1)\]

\[8)\ 0,09t^{4} - t^{6} =\]

\[= t^{4}\left( 0,09 - t^{2} \right) =\]

\[= t^{4}(0,3 - t)(0,3 + t)\]

\[9)\frac{16}{49}a^{2}b^{4}c^{5} - b^{2}c^{3} =\]

\[= b^{2}c^{3}\left( \frac{16}{49}a^{2}b^{2}c^{2} - 1 \right) =\]

\[= b^{2}c^{3}\left( \frac{4}{7}abc - 1 \right)\left( \frac{4}{7}abc + 1 \right)\]

\[\boxed{\text{707.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1){\ \left( 10a^{2} - 7ab^{2} \right)}^{2} =\]

\[= \left( 10a^{2} \right)^{2} - 2 \cdot 10a^{2} \cdot 7ab^{2} =\]

\[= 100a^{4} - 140a^{3}b^{2} + 49a^{2}b^{4}\]

\[2)\ \left( 0,8b^{3} + 0,2b^{2}c^{4} \right)^{2} =\]

\[= 0,64b^{6} + 0,32b^{5}c^{4} + 0,04b^{4}c^{8}\]

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

\[= \left( \frac{4}{3}a^{2}b + \frac{9}{4}ab^{2} \right)^{2} =\]

\[= \frac{16}{9}a^{4}b^{2} + 6a^{3}b^{3} + \frac{81}{16}a^{2}b^{4} =\]

\[= 1\frac{7}{9}a^{4}b^{2} + 6a^{3}b^{3} + 5\frac{1}{16}a^{2}b^{4}\]

\[4)\ \left( 2\frac{1}{3}x^{3}y^{2} - \frac{9}{14}y^{8}x \right)^{2} =\]

\[= \left( \frac{7}{3}x^{3}y^{2} - \frac{9}{14}y^{8}x \right)^{2} =\]