Решебник по алгебре 7 класс Мерзляк ФГОС Задание 657

 

\[\boxed{\text{657\ (657).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ x² + y² + 8x - 10y + 41 = 0\]

\[\left( x^{2} + 8x + 16 \right) + \left( y^{2} - 10y + 25 \right) =\]

\[= 0\]

\[(x + 4)^{2} + (y - 5)^{2} = 0\]

\[x = - 4,\ \ y = 5\]

\[Ответ:при\ \ x = - 4,\ y = 5.\]

\[2)\ x² + 37y² + 12xy - 2y + 1 =\]

\[= 0\]

\[\left( x^{2} + 12xy + 36y^{2} \right) + \left( y^{2} - 2y + 1 \right) =\]

\[= 0\]

\[(x + 6y)^{2} + (y - 1)^{2} = 0\]

\[y = 1,\ \ x = - 6y = - 6 \cdot 1 =\]

\[= - 6\]

\[Ответ:при\ x = - 6,\ y = 1.\]

\[\boxed{\text{657.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ 16 - b^{2} = (4 - b)(4 + b)\]

\[2)\ c^{2} - 49 = (c - 7)(c + 7)\]

\[3)\ 0,04 - a^{2} =\]

\[= (0,2 - a)(0,2 + a)\]

\[4)\ x^{2} - \frac{4}{9} = \left( x - \frac{2}{3} \right)\left( x + \frac{2}{3} \right)\]

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

\[6)\ 81c^{2} - 64d^{2} =\]

\[= (9c - 8d)(9c + 8d)\]

\[7)\ 0,09x^{2} - 0,25y^{2} =\]

\[= (0,3x - 0,5y)(0,3x + 0,5y)\]

\[8)\ a^{2}b^{4} - c^{6}d^{8} =\]

\[= (ab^{2} - c^{3}d^{4})(ab^{2} + c^{3}d^{4})\]

\[9)\ 4a^{2}c^{2} - 9x^{2}y^{2} =\]

\[= (2ac - 3xy)(2ac + 3xy)\]

\[10)\ x^{24} - y^{22} =\]

\[= (x^{12} - y^{11})(x^{12} + y^{11})\]

\[11) - 1600 + a^{12} =\]

\[= a^{12} - 1600 =\]

\[= (a^{6} - 40)(a^{6} + 40)\]

\[12)\ a^{18} - \frac{49}{64} =\]

\[= \left( a^{9} - \frac{7}{8} \right)\left( a^{9} + \frac{7}{8} \right)\]