Решебник по алгебре 8 класс Мерзляк ФГОС Задание 60

 

\[\boxed{\text{60}\text{\ (60)}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ ax = 1\]

\[если\ a = 0 \Longrightarrow 0 \cdot x =\]

\[= 1 - корней\ нет.\]

\[если\ a \neq 0 \Longrightarrow \ x = \frac{1}{a}.\]

\[2)\ ax = a\]

\[если\ a = 0 \Longrightarrow \ \]

\[\Longrightarrow x - любое\ число;.\]

\[если\ \ a \neq 0 \Longrightarrow \ x = \frac{a}{a} = 1.\]

\[3)\ (a - 6)x = a^{2} - 12a + 36\]

\[(a - 6)x = (a - 6)^{2}\]

\[если\ a = 6 \Longrightarrow 0 \cdot x = 0;\ \ \]

\[x - любое\ число.\]

\[если\ a \neq 6 \Longrightarrow x = \frac{(a - 6)^{2}}{(a - 6)};\ \ \]

\[x = a - 6.\]

\[4)\ \left( a^{2} - 4 \right) \cdot x = a - 2\]

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

\[если\ a = - 2:\]

\[- 4 \cdot 0 \cdot x = - 4\]

\[0 \cdot x = 1 \Longrightarrow корней\ нет.\]

\[если\ a = 2:\]

\[0 \cdot 4 \cdot x = 0\]

\[0 \cdot x = 0 \Longrightarrow \ \ x - любое\ число.\]

\[если\ a \neq 2\ \ и\ a \neq - 2:\ \]

\[x = \frac{(a - 2)}{(a - 2)(a + 2)} \Longrightarrow \ \ x =\]

\[= \frac{1}{a + 2}.\]

\[\boxed{\text{60.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ (x + 2)(x - 9) - 3x(3 - 2x) =\]

\[= x^{2} - 9x + 2x - 18 - 9x + 6x^{2} =\]

\[= 7x^{2} - 16x - 18\]

\[2)\ (a + 5)(a - 2) +\]

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

\[= a^{2} - 2a + 5a - 10 + a^{2} -\]

\[- 5a + 4a - 20 = 2a^{2} + 2a - 30\]

\[3)\ (y - 8)(2y + 1) -\]

\[- (3y + 1)(y - 6) =\]

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

\[+ 18y - y + 6 = - y^{2} + 2y - 2\]

\[4)\ (2x - 3y)(2x + 3y) +\]

\[+ (3x + 2y)(3x - 2y) =\]

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

\[= 13x^{2} - 13y^{2}\]

\[5)\ (x + 1)^{2} - (x - 3)(x + 3) =\]

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

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

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

\[= 2x + 10\]

\[6)\ (y - 4)(y + 3) - (y - 6)^{2} =\]

\[= y^{2} + 3y - 4y - 12 -\]

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

\[= y^{2} + 3y - 4y - 12 - y^{2} +\]

\[+ 12y - 36 = 11y - 48\]