Решебник по алгебре 8 класс Мерзляк ФГОС Задание 564

 

\[\boxed{\mathbf{564\ \ (564).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{- m^{19}} = \left| m^{9} \right|\sqrt{- m} =\]

\[= {- m}^{9}\sqrt{- m}\]

\[2)\ \sqrt{a^{23}b^{24}} = a^{11}b^{12}\sqrt{a}\]

\[3)\ \sqrt{49a^{2}b} = 7|a|\sqrt{b} = - 7a\sqrt{b};\ \ \]

\[a < 0\]

\[4)\ \sqrt{a^{9}b^{9}} = a^{4}b^{4}\sqrt{\text{ab}}\]

\[5)\ \sqrt{27x^{15}y^{34}} = - 3x^{7}y^{17}\sqrt{3x};\ \ \]

\[y < 0\]

\[6)\ \sqrt{- 50m^{6}n^{6}p^{7}} =\]

\[= - 5m^{3}n^{3}p^{3}\sqrt{- 2p};\ \ m > 0,\]

\[n > 0\]

\[\boxed{\mathbf{5}\mathbf{6}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( 2 - \sqrt{3} \right)\left( \sqrt{3} + 1 \right) =\]

\[= 2\sqrt{3} + 2 - 3 - \sqrt{3} = \sqrt{3} - 1;\]

\[2)\ \left( \sqrt{2} + \sqrt{5} \right)\left( 2\sqrt{2} - \sqrt{5} \right) =\]

\[= 2 \cdot 2 - \sqrt{10} + 2\sqrt{10} - 5 =\]

\[= \sqrt{10} - 1;\]

\[3)\ \left( a + \sqrt{b} \right)\left( a - \sqrt{b} \right) = a² - b;\]

\[4)\ \left( \sqrt{b} - \sqrt{c} \right)\left( \sqrt{b} + \sqrt{c} \right) = b - c;\]

\[5)\ \left( 4 + \sqrt{3} \right)\left( 4 - \sqrt{3} \right) = 16 - 3 =\]

\[= 13;\]

\[6)\ \left( y - \sqrt{7} \right)\left( y + \sqrt{7} \right) = y² - 7;\]

\[7)\ \left( 4\sqrt{2} - 2\sqrt{3} \right)\left( 2\sqrt{3} + 4\sqrt{2} \right) =\]

\[= \left( 4\sqrt{2} \right)^{2} - \left( 2\sqrt{3} \right)^{2} = 32 - 12 =\]

\[= 20;\]

\[8)\ \left( m + \sqrt{n} \right)^{2} = m² + 2m\sqrt{n} + n;\]

\[9)\ \left( \sqrt{a} - \sqrt{b} \right)^{2} = a - 2\sqrt{\text{ab}} + b;\]

\[10)\ \left( 2 - 3\sqrt{3} \right)^{2} =\]

\[= 4 - 12\sqrt{3} + 27 = 31 - 12\sqrt{3}.\]