Решебник по алгебре 7 класс Мерзляк Проверь себя №5

\[\boxed{\text{Задание}\text{\ }\text{№}\text{5\ }\text{«Проверьте}\text{\ }\text{себя»}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\boxed{\text{1.}}\]

\[(x - 6)\left( x^{2} + 6x + 36 \right) =\]

\[= x^{3} - 216\]

\[Ответ:\ \ В).\]

\[\boxed{\text{2.}}\]

\[y^{3} - 64 = (y - 4) \cdot M\]

\[y^{3} - 64 =\]

\[= (y - 4)\left( y^{2} + 4y + 16 \right)\]

\[M = y^{2} + 4y + 16\]

\[Ответ:\ Г).\]

\[\boxed{\text{3.}}\]

\[\left( a^{2} + 2b^{3} \right)\left( a^{4} - 2a^{2}b^{3} + 4b^{6} \right) =\]

\[= a^{6} + 8b^{9}\]

\[Ответ:\ Г).\]

\[\boxed{\text{4.}}\]

\[3c^{2} - 48 = 3 \cdot \left( c^{2} - 16 \right) =\]

\[= 3 \cdot (c - 4)(c + 4)\]

\[Ответ:Б).\]

\[\boxed{\text{5.}}\]

\[7a^{2} - 42a + 63 =\]

\[= 7 \cdot \left( a^{2} - 6a + 9 \right) =\]

\[= 7 \cdot (a - 3)^{2}\]

\[Ответ:Б).\]

\[\boxed{\text{6.}}\]

\[a^{8} - a^{6} = a^{6} \cdot \left( a^{2} - 1 \right) =\]

\[= a^{6}(a - 1)(a + 1)\]

\[Ответ:Б).\]

\[\boxed{\text{7.}}\]

\[m^{2} - n^{2} + m + n =\]

\[= (m - n)(m + n) + (m + n) =\]

\[= (m + n)(m - n + 1)\]

\[Ответ:А).\]

\[\boxed{\text{8.}}\]

\[x^{2} - y^{2} + 14y - 49 =\]

\[= x^{2} - (y - 7)^{2} =\]

\[= (x - y + 7)(x + y - 7)\]

\[Ответ:В).\]

\[\boxed{\text{9.}}\]

\[81a^{4} - 1 =\]

\[= \left( 9a^{2} - 1 \right)\left( 9a^{2} + 1 \right) =\]

\[= (3a - 1)(3a + 1)\left( 9a^{2} + 1 \right)\]

\[Ответ:А).\]

\[\boxed{\text{10.}}\]

\[49x - x^{2} = 0\]

\[x \cdot (49 - x) = 0\]

\[x = 0\ \ \ или\ \ x = 49\]

\[Ответ:В).\]

\[\boxed{\text{11.}}\]

\[x^{3} + 3x^{2} - x - 3 = 0\]

\[x^{2} \cdot (x + 3) - (x + 3) = 0\]

\[(x + 3)\left( x^{2} - 1 \right) = 0\]

\[(x + 3)(x - 1)(x + 1) = 0\]

\[x = - 3,\ \ x = 1,\ \ x = - 1\]

\[Ответ:Г).\]

\[\boxed{\text{12.}}\]

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

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

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

\[= (x - 2)^{2} \cdot (x + 2)^{2}\]

\[Ответ:Б).\]