ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 889

\[\boxed{\text{889.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ (x - 2)(x + 2) - x(x + 5) = x^{2} - 4 - x^{2} - 5x = - 5x - 4\]

\[\textbf{б)}\ m(m - 4) + (3 - m)(3 + m) = m² - 4m + 9 - m^{2} = - 4m + 9\]

\[\textbf{в)}\ (4x - a)(4x + a) + 2x(x - a) = 16x² - a^{2} + 2x^{2} - 2xa =\]

\[= 18x^{2} - 2ax - a^{2}\]

\[\textbf{г)}\ 2a(a + b) - (2a + b)(2a - b) = 2a^{2} + 2ab - \left( 4a^{2} - b^{2} \right) =\]

\[= 2a² + 2ab - 4a^{2} + b^{2} = - 2a^{2} + 2ab + b²\]

\[\textbf{д)}\ (5a - 3c)(5a + 3c) - (7c - a)(7c + a) = 25a^{2} - 9c^{2} - \left( 49c^{2} - a^{2} \right) =\]

\[= 25a² - 9c^{2} - 49c^{2} + a^{2} = 26a^{2} - 58c²\]

\[\textbf{е)}\ (4b + 10c)(10c - 4b) + ( - 5c + 2b)(5c + 2b) =\]

\[= 100c² - 16b^{2} + 4b^{2} - 25c^{2} = 75c^{2} - 12b²\]

\[\textbf{ж)}\ (3x - 4y)^{2} - (3x - 4y)(3x + 4y) = 9x^{2} - 24xy + 16y^{2} - \left( 9x^{2} - 16y^{2} \right) =\]

\[= 9x^{2} - 24xy + 16y^{2} - 9x^{2} + 16y^{2} = 32y^{2} - 24xy\]

\[\textbf{з)}\ (2a + 6b)(6b - 2a) - (2a + 6b)^{2} = 36b^{2} - 4a^{2} - \left( 4a^{2} + 24ab + 36b^{2} \right) =\]

\[= 36b² - 4a^{2} - 4a^{2} - 24ab - 36b^{2} = - 8a^{2} - 24ab\]