Решебник по алгебре 8 класс Мерзляк ФГОС Задание 874

 

\[\boxed{\mathbf{874\ (874).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\text{\ a}^{- 7} \cdot a^{10} = a^{3}\]

\[2)\ a^{- 9} \cdot a^{5} = a^{- 4}\]

\[3)\ a^{17} \cdot a^{- 4} \cdot a^{- 11} = a^{2}\]

\[4)\ a^{- 2}\ :a^{3} = a^{- 5}\]

\[5)\ a^{12}\ :a^{- 4} = a^{16}\]

\[6)\ a^{- 7}\ :a^{- 11} = a^{4}\]

\[7)\ a^{- 12}\ :a^{- 10} \cdot a^{4} = a^{2}\]

\[8)\ \left( a^{3} \right)^{- 5} = a^{- 15}\]

\[9)\ \left( a^{- 12} \right)^{- 2} = a^{24}\]

\[10)\ \left( a^{- 3} \right)^{4}\ :\left( a^{- 2} \right)^{5}\ :\left( a^{- 1} \right)^{- 7} =\]

\[= a^{- 12}\ :a^{- 10}\ :a^{7} = a^{- 9}\]

\[11)\ \left( m^{- 3}n^{4}p^{7} \right)^{- 4} = m^{12}n^{- 16}p^{- 28}\]

\[12)\ \left( a^{- 1}b^{- 2} \right)^{- 3} = a^{3}b^{6}\]

\[13)\ \left( x^{3}y^{- 4} \right)^{5} \cdot \left( x^{- 2}y^{- 3} \right)^{3} =\]

\[= x^{15}y^{- 20} \cdot x^{- 6}y^{- 9} = x^{9}y^{- 29}\]

\[14)\ \left( \frac{a^{11}b^{- 7}}{c^{- 3}d^{4}} \right)^{- 3} = \frac{a^{- 33}b^{21}}{c^{9}d^{- 12}} =\]

\[= a^{- 33}b^{21}c^{- 9}d^{12}\]

\[15)\ \left( \frac{a^{- 7}}{b^{5}} \right)^{- 3} \cdot \left( \frac{a^{4}}{b^{- 7}} \right)^{- 5} =\]

\[= \frac{a^{21}}{b^{- 15}} \cdot \frac{a^{- 20}}{b^{35}} = \frac{a}{b^{20}} = ab^{- 20}\]

\[\boxed{\mathbf{8}\mathbf{74}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ (p - 3)(p + 4) < p(p + 1)\]

\[p^{2} - 3p + 4p - 12 < p^{2} + p\]

\[p - p < 12\]

\[0p < 12\]

\[чтд.\]

\[2)\ (x + 1)^{2} > x(x + 2)\]

\[x^{2} + 2x + 1 > x^{2} + 2x\]

\[1 > 0\]

\[чтд.\]

\[3)\ (a - 5)(a + 2) > (a + 5)(a - 8)\]

\[a^{2} - 5a + 2a - 10 > a^{2} + 5a - 8a - 40\]

\[- 3a + 3a > - 40 + 10\]

\[0a > - 30\]

\[чтд.\]

\[4)\ y(y + 8) < (y + 4)^{2}\ \]

\[y^{2} + 8y < y^{2} + 8y + 16\]

\[0y < 16\]

\[чтд.\]

\[5)\ (2a - 5)^{2} \leq 6a^{2} - 20a + 25\]

\[4a^{2} - 20a + 25 - 6a^{2} + 20a - 25 \leq 0\]

\[- 2a^{2} \leq 0\]

\[чтд.\]

\[6)\ a^{2} + 4 \geq 4a\]

\[a^{2} - 4a + 4 \geq 0\]

\[(a - 2)^{2} \geq 0\]

\[чтд.\]