ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 998

\[\boxed{\text{998.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ (x - 5)^{2} - 16 =\]

\[= (x - 5 - 4)(x - 5 + 4) =\]

\[= (x - 9)(x - 1)\]

\[\textbf{б)}\ (b + 7)^{2} - 9 =\]

\[= (b + 7 - 3)(b + 7 + 3) =\]

\[= (b + 4)(b + 10)\]

\[\textbf{в)}\ 25 - (3 - x)^{2} =\]

\[= (5 - 3 + x)(5 + 3 - x) =\]

\[= (2 + x)(8 - x)\]

\[\textbf{г)}\ 81 - (a + 7)^{2} =\]

\[= (9 - a - 7)(9 + a + 7) =\]

\[= (2 - a)(16 + a)\]

\[\textbf{д)}\ (7x - 4)^{2} - (2x + 1)^{2} =\]

\[= (7x - 4 - 2x - 1)\]

\[(7x - 4 + 2x + 1) =\]

\[= (5x - 5)(9x - 3) =\]

\[= 15 \cdot (x - 1)(3x - 1)\]

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

\[= (n - 2 - 3n - 1)\]

\[(n - 2 + 3n + 1) =\]

\[= ( - 2n - 3)(4n - 1)\]

\[\textbf{ж)}\ 9 \cdot (a + 1)^{2} - 1 =\]

\[= \left( 3 \cdot (a + 1) - 1 \right)\]

\[\left( 3 \cdot (a + 1) + 1 \right) =\]

\[= (3a + 3 - 1)(3a + 3 + 1) =\]

\[= (3a + 2)(3a + 4)\]

\[\textbf{з)}\ 4 - 25 \cdot (x - 3)^{2} =\]

\[= \left( 2 - 5 \cdot (x - 3) \right) =\]

\[= \left( 2 + 5 \cdot (x - 3) \right) =\]

\[= (2 - 5x + 15)(2 + 5x - 15) =\]

\[= (17 - 5x)(5x - 13)\ \]