Решебник по алгебре 8 класс Мерзляк ФГОС Задание 389

 

\[\boxed{\text{389\ (389).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \sqrt{16 + 9} = \sqrt{25} = 5\]

\[2)\ \sqrt{16} + \sqrt{9} = 4 + 3 = 7\]

\[3)\ \sqrt{36} - \sqrt{49} = 6 - 7 = - 1\]

\[4)\ \sqrt{36} \cdot \sqrt{49} = 6 \cdot 7 = 42\]

\[5)\ 5\sqrt{4} - \sqrt{25} = 5 \cdot 2 - 5 = 5\]

\[6)\ \sqrt{0,81} + \sqrt{0,01} = 0,9 + 0,1 = 1\]

\[7)\frac{1}{3}\sqrt{0,09} - 2 = \frac{0,3}{3} - 2 = - 1,9\]

\[8) - 2\sqrt{0,16} + 0,7 =\]

\[= - 2 \cdot 0,4 + 0,7 =\]

\[= - 0,8 + 1,7 = - 0,1\]

\[9)\ \left( \sqrt{13} \right)^{2} - 3 \cdot \left( \sqrt{8} \right)^{2} =\]

\[= 13 - 3 \cdot 8 = 13 - 24 = - 11\]

\[10)\frac{1}{6} \cdot \left( \sqrt{18} \right)^{2} - \left( \frac{1}{2}\sqrt{24} \right)^{2} =\]

\[= \frac{1}{6} \cdot 18 - \frac{1}{4} \cdot 24 = \frac{18}{6} - \frac{24}{4} =\]

\[= 3 - 6 = - 3\]

\[11)\ 50 \cdot \left( - \frac{1}{5}\sqrt{2} \right)^{2} = 50 \cdot \frac{1}{25} \cdot 2 =\]

\[= \frac{50 \cdot 2}{25} = 4\]

\[12)\ \sqrt{4 \cdot 5^{2} - 6^{2}} = \sqrt{4 \cdot 25 - 36} =\]

\[= \sqrt{100 - 36} = \sqrt{64} = 8\]

\[\boxed{\text{3}\text{89}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{1^{\backslash b}}{6a} - \frac{6a + b}{6ab} = \frac{b - 6a - b}{6ab} =\]

\[= - \frac{6a}{6ab} = - \frac{1}{b}\]

\[2)\ \frac{5}{a^{2} - 4b^{2}} - \frac{5}{a - 2b} =\]

\[= \frac{5}{(a - 2b)(a + 2b)} - \frac{5^{\backslash a + 2b}}{a - 2b} =\]

\[= \frac{5 - 5a - 10b}{(a - 2b)(a + 2b)} =\]

\[= \frac{5 - 5a - 10b}{a^{2} - 4b}\]

\[3)\ \frac{2m}{m^{2} - 36n^{2}} \cdot \left( mn - 6n^{2} \right) =\]

\[= \frac{2m \cdot n(m - 6n)}{(m - 6n)(m + 6n)} = \frac{2mn}{m + 6n}\]

\[4)\ \frac{7c^{2}}{c^{2} - 16a^{2}}\ :\frac{c}{c^{2} - 4ac} =\]

\[= \frac{7c^{2}}{(c - 4a)(c + 4a)} \cdot \frac{c(c - 4a)}{c} =\]

\[= \frac{7c^{2}}{c + 4a}\]