\[\boxed{\text{486\ (486).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ a^{n + 1} + a^{n} + a + 1 =\]
\[= a^{n} \cdot (a + 1) + (a + 1) =\]
\[= (a + 1) \cdot \left( a^{n} + 1 \right).\]
\[2)\ b^{n + 2} - b - 1 + b^{n + 1} =\]
\[= \left( b^{n + 2} + b^{n + 1} \right) - (b + 1) =\]
\[= b^{n + 1} \cdot (b + 1) - (b + 1) =\]
\[= (b + 1) \cdot \left( b^{n + 1} - 1 \right).\]
\[3)\ 3y^{n + 3} - 3y^{2} - 5 + 5y^{n + 1} =\]
\[= \left( 3y^{2} + 5 \right) \cdot \left( y^{n + 1} - 1 \right)\text{.\ }\]
\[\boxed{\text{486.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \frac{2x + 3}{3} - \frac{5x + 13}{6} + \frac{5 - 2x}{2} = 6\ \ | \cdot 6\]
\[2(2x + 3) - 5x - 13 + 3(5 - 2x) = 36\]
\[4x + 6 - 5x - 13 + 15 - 6x = 36\]
\[- 7x = 36 - 8\]
\[- 7x = 28\]
\[x = - 4.\]
\[2)\ \frac{4x^{2} + 5x}{14} + \frac{10 - 2x^{2}}{7} = 5\ \ \ | \cdot 14\]
\[4x^{2} + 5x + 2\left( 10 - 2x^{2} \right) = 70\]
\[4x^{2} + 5x - 20 - 4x^{2} = 70\]
\[5x = 90\]
\[x = \frac{90}{5} = 18.\]