Решебник по алгебре 7 класс Мерзляк Проверь себя №3

\[\boxed{\text{Задание}\text{\ }\text{№}\text{3\ }\text{«Проверьте}\text{\ }\text{себя»}\text{.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[3y^{2}\left( y^{3} + 1 \right) =\]

\[= 3y^{2} \cdot y^{3} + 3y^{2} \cdot 1 =\]

\[= 3y^{5} + 3y^{2}\]

\[Ответ:\ \ Г).\]

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

\[- 9y \cdot (y - 3) + 4,5y \cdot (2y - 4) =\]

\[Ответ:Г).\ \]

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

\[(x - 3) \cdot (x + 7) =\]

\[= x^{2} + 7x - 3x - 21 =\]

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

\[Ответ:А).\ \]

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

\[(3x + 2)(2x - 1) - (5x - 2)(x - 4) =\]

\[= x^{2} + 23x - 10\]

\[Ответ:\ Б).\]

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

\[3mn - 4mk = m(3n - 4k)\]

\[Ответ:\ Б).\]

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

\[m^{2}n + mn^{2} = mn(m + n)\]

\[Ответ:В).\ \]

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

\[mn - mn^{2} = mn(1 - n)\]

\[Ответ:А).\ \]

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

\[2x^{2} - 4x^{6} = 2x^{2}\left( 1 - 2x^{4} \right)\]

\[Ответ:\ \ Б).\]

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

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

\[x \cdot (x - 2) = 0\]

\[x = 0\ \ \ или\ \ \ x - 2 = 0\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 2\]

\[Ответ:\ В).\]

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

\[ax - ay + 5x - 5y =\]

\[= a \cdot (x - y) + 5 \cdot (x - y) =\]

\[= (x - y)(a + 5)\]

\[Ответ:\ А).\]

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

\[\frac{x - 1}{2} - \frac{x + 1}{3} = 1\ \ \ \ \ \ \ | \cdot 6\]

\[3 \cdot (x - 1) - 2 \cdot (x + 1) = 1 \cdot 6\]

\[3x - 3 - 2x - 2 = 6\]

\[x - 5 = 6\]

\[x = 6 + 5\]

\[x = 11\]

\[Ответ:\ А).\]

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

\[a^{2} - 7a + 3 = 2\]

\[2a^{2} - 14a + 10 =\]

\[= 2 \cdot \left( \underset{= 2}{\overset{a^{2} - 7a + 3}{︸}} + 2 \right) =\]

\[= 2 \cdot 4 = 8\]

\[Ответ:В).\ \]