ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 991

\[\boxed{\text{991.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ (x - 5)^{2} + 2x(x - 3) = x² -\]

\[- 10x + 25 + 2x^{2} - 6x =\]

\[= 3x^{2} - 16x + 25\]

\[\textbf{б)}\ (y + 8)^{2} - 4y(y - 2) = y^{2} +\]

\[16y + 64 - 4y^{2} + 8y = - 3y^{2} +\]

\[+ 24y + 64\]

\[\textbf{в)}\ (a - 4)(a + 4) + (2a - 1)^{2} =\]

\[= a^{2} - 16 + 4a^{2} - 4a + 1 =\]

\[= 5a^{2} - 4a - 15\]

\[\textbf{г)}\ (b - 3)(b + 3) - (b + 2)^{2} =\]

\[= b^{2} - 9 - \left( b^{2} + 4b + 4 \right) =\]

\[= b^{2} - 9 - b^{2} - 4b - 4 =\]

\[= - 4b - 13\]

\[\textbf{д)}\ (2a - 5)^{2} - (5a - 2)^{2} =\]

\[= 4a^{2} - 20a + 25 -\]

\[- \left( 25a^{2} - 20a + 4 \right) =\]

\[= 4a^{2} - 20a + 25 - 25a^{2} +\]

\[+ 20a - 4 = - 21a^{2} + 21\]

\[\textbf{е)}\ (3b - 1)^{2} + (1 - 3b)^{2} =\]

\[= 9b^{2} - 6b + 1 + 1 - 6b +\]

\[+ 9b^{2} = 18b^{2} - 12b + 2\]

\[\textbf{ж)}\ (2x + 1)^{2} - (x + 7)(x - 3) =\]

\[= 4x^{2} + 4x + 1 -\]

\[- \left( x^{2} - 3x + 7x - 21 \right) =\]

\[= 4x^{2} + 4x + 1 - x^{2} - 4x +\]

\[+ 21 = 3x^{2} + 22\]

\[\textbf{з)}\ (3y - 2)^{2} - (y - 9)(9 - y) =\]

\[= 9y^{2} - 12y + 4 -\]

\[- \left( 9y - y^{2} - 81 + 9y \right) =\]

\[= 9y^{2} - 12y + 4 - 9y + y^{2} +\]

\[+ 81 - 9y = 10y^{2} - 30y + 85\ \]