Решебник по алгебре 8 класс Мерзляк ФГОС Задание 539

 

\[\boxed{\mathbf{539\ (539).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( \sqrt{7} + 3 \right)\left( 3\sqrt{7} - 1 \right) =\]

\[= 3 \cdot 7 - \sqrt{7} + 9\sqrt{7} - 3 =\]

\[= 8\sqrt{7} + 18;\]

\[2)\ \left( 4\sqrt{2} - \sqrt{3} \right)\left( 2\sqrt{2} + 5\sqrt{3} \right) =\]

\[= 16 + 20\sqrt{6} - 2\sqrt{6} - 15 =\]

\[= 18\sqrt{6} + 1;\]

\[3)\ \left( \sqrt{p} - q \right)\sqrt{p} + q) = p - q^{2};\]

\[4)\ \left( 6 - \sqrt{13} \right)\left( 6 + \sqrt{13} \right) =\]

\[= 36 - 13 = 23;\]

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

\[6)\ \left( \sqrt{19} + \sqrt{17} \right)\left( \sqrt{19} - \sqrt{17} \right) =\]

\[= 19 - 17 = 2;\]

\[7)\ \left( \sqrt{6} + \sqrt{2} \right)^{2} = 6 + 2\sqrt{12} + 2 =\]

\[= 4\sqrt{3} + 8;\]

\[8)\ \left( 3 - 2\sqrt{15} \right)² =\]

\[= 9 - 12\sqrt{15} + 60 =\]

\[= 69 - 12\sqrt{15}.\]

\[\boxed{\mathbf{5}\mathbf{3}\mathbf{9}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{9a^{16}} = 3a^{8};\]

\[2)\ \sqrt{0,81d^{6}} = 0,9d^{3};\ \ \ \]

\[если\ d \geq 0;\]

\[3) - 5\sqrt{4x^{2}} = 10x;\ \ если\ x \leq 0;\]

\[4) - 0,1\sqrt{100z^{10}} =\]

\[= - 0,1 \cdot 10\left| z^{5} \right| = - z^{5};\ \ \]

\[если\ z \geq 0;\]

\[5)\ \sqrt{p^{6}q^{8\ \ }} = \left| p^{3} \right|q^{4} = p^{3}q^{4};\ \ \]

\[если\ p \geq 0;\]

\[6)\ \sqrt{25m^{34}n^{38}} = 5\left| m^{17} \right|\left| n^{19} \right| =\]

\[= 5m^{17}n^{19};\ \ если\ m \leq 0,\ n \leq 0;\]

\[7)\ ab^{2}\sqrt{a^{4}b^{18}c^{22}} =\]

\[= ab^{2}a^{2\left| b^{9} \right|\left| c^{11} \right|} = - a^{3}b^{11}c^{11};\ \ \]

\[если\ b \geq 0,\ c \leq 0;\]

\[8) - \frac{8m^{3}p^{4}}{k^{2}} \cdot \sqrt{\frac{625k^{30}p^{40}}{144m^{6}}} =\]

\[= \frac{- 8m^{3}p^{4} \cdot 25k^{15}p^{20}}{- k^{2} \cdot 12m^{3}} =\]

\[= \frac{50p^{24}k^{13}}{3};\ \]

\[если\ m < 0;\ \ k > 0.\]