Решебник по алгебре 7 класс Мерзляк Задание 536

\[\boxed{\text{536\ (536).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ b^{2} - d^{2} = (b - d)(b + d)\]

\[2)\ x^{2} - 1 = (x - 1)(x + 1)\]

\[3) - x^{2} + 1 = 1 - x^{2} =\]

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

\[4)\ 36 - c^{2} = (6 - c)(6 + c)\]

\[5)\ 4 - 25a^{2} = (2 - 5a)(2 + 5a)\]

\[6)\ 49a^{2} - 100 =\]

\[= (7a - 10)(7a + 10)\]

\[7)\ 900 - 81k^{2} =\]

\[= (30 - 9k)(30 + 9k)\]

\[8)\ 16x^{2} - 121y^{2} =\]

\[= (4x - 11y)(4x + 11y)\]

\[9)\ b^{2}c^{2} - 1 = (bc - 1)(bc + 1)\]

\[10)\ \frac{1}{4}x^{2} - \frac{1}{9}y^{2} =\]

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

\[11) - 4a^{2}b^{2} + 25 =\]

\[= 25 - 4a^{2}b^{2} =\]

\[= (5 - 2ab)(5 + 2ab)\]

\[12)\ 144x^{2}y^{2} - 400 =\]

\[= (12xy - 20)(12xy + 20)\]

\[13)\ a^{2}b^{2}c^{2} - 1 =\]

\[= (abc - 1)(abc + 1)\]

\[14)\ 100a^{2} - 0,01b^{2} =\]

\[= (10a - 0,1b)(10a + 0,1b)\]

\[15)\ a^{4} - b^{2} = (a^{2} - b)(a^{2} + b)\]

\[16)\ p^{2}t^{2} - 0,36k^{2}d^{2} =\]

\[= (pt - 0,6kd)(pt + 0,6kd)\]

\[17)\ y^{10} - 9 = (y^{5} - 3)(y^{5} + 3)\]

\[18)\ 4x^{12} - 1\frac{11}{25}y^{16} =\]

\[= 4x^{12} - \frac{36}{25}y^{16} =\]

\[= \left( 2x^{6} - \frac{6}{5}y^{8} \right)\left( 2x^{6} + \frac{6}{5}y^{8} \right)\]