Решебник по алгебре 8 класс Мерзляк ФГОС Задание 895

 

\[\boxed{\mathbf{895\ (895).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{q^{2}} = |q| = q,\ \ q > 0\]

\[2)\ \sqrt{t^{2}} = |t| = - t,\ \ t \leq 0\]

\[3)\ \sqrt{49m^{2}n^{8}} = 7|m|n^{4} = 7mn^{4},\]

\[m \geq 0\]

\[4)\ \sqrt{0,81a^{6}b^{10}} = 0,9\left| a^{3} \right||b^{5}| =\]

\[= - 0,9a^{3}b^{5};\ \ \ \ \ a \geq 0,\ \ b \leq 0\]

\[5)\ \frac{1}{5}x\sqrt{100x^{26}} = \frac{1}{5}x \cdot 10|x^{13}| =\]

\[= - 2x^{14},\ \ x \leq 0\]

\[6)\ \frac{\sqrt{a^{6}b^{20}c^{34}}}{ab^{8}c^{12}} = \frac{\left| a^{3} \right|b^{10}\left| c^{17} \right|}{ab^{8}c^{12}} =\]

\[= - a^{2}b^{2}c^{5},\ \ a > 0,\ \ c < 0\]

\[7)\ \frac{1,2x^{3}}{y^{5}} \cdot \sqrt{\frac{y^{14}}{x^{10}}} = \frac{1,2x^{3} \cdot \left| y^{7} \right|}{y^{5} \cdot {|x}^{5}|} =\]

\[= \frac{- 1,2y²}{x²},\ \ y > 0,\ \ x < 0\]

\[8) - 0,1x^{2}\sqrt{1,96x^{18}y^{16}} =\]

\[= - 0,1x^{2} \cdot 1,4\left| x^{9} \right|y^{8} =\]

\[= 0,14x^{11}y^{8},\ \ x \leq 0\]

\[\boxed{\mathbf{8}\mathbf{95}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 2a(5a - 7) - 5a(3 - 2a) =\]

\[= 10a^{2} - 14a - 15a + 10a^{2} =\]

\[= 20a^{2} - 29a\]

\[2)\ (2b - 3)(4b + 9) = 8b^{2} +\]

\[+ 18b - 12b - 27 =\]

\[= 8b^{2} + 6b - 27\]

\[3)\ (2c - 6)(8c + 5) -\]

\[- (5c + 2)(5c - 2) =\]

\[= 16c^{2} + 10c - 48c - 30 -\]

\[- 25c^{2} + 4 = - 9c^{2} - 38c - 26\]

\[4)\ 16m^{2} - (3 - 4m)(3 + 4m) =\]

\[= 16m^{2} - 9 + 16m^{2} = 32m^{2} - 9\]

\[5)\ (2x - 1)^{2} + (2x + 1)^{2} = 4x^{2} -\]

\[- 4x + 1 + 4x^{2} + 4x + 1 =\]

\[= 8x^{2} + 2\]

\[6)\ (x - 4)(x + 4) - (x - 8)^{2} =\]

\[= x^{2} - 16 - x^{2} + 16x - 64 =\]

\[= 16x - 80\ \]