Предмет: Алгебра
Автор: gricundasa
Вопрос задан 2 года назад

Помогите пожайлуста! Очень срочно
Даю 60 баллов

Приложения:

Ответы

Ответ дал: NNNLLL54

Ответ:

$1a)\ \ \sqrt{16\cdot 81}=4\cdot 9=36\ \ ,\ \ \sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\ \ ,\ \ \sqrt{\dfrac{9}{49}\cdot \dfrac{1}{121}}=\dfrac{3}{7\cdot 11}=\dfrac{3}{77}\\\\\\b)\ \ \sqrt{50}\cdot \sqrt2=\sqrt{100}=10\ \ ,\ \ \ \dfrac{\sqrt{343}}{\sqrt7}=\sqrt{49}=7\\\\c)\ \ \sqrt{162\cdot 50}=\sqrt{81\cdot 100}=9\cdot 10=90\\\\\sqrt{0,8\cdot 14,4\cdot 0,5}=\sqrt{0,4\cdot (1,44\cdot 10)}=\sqrt{(0,04\cdot 10)\cdot (1,44\cdot 10)}=\\\\=0,2\cdot 1,2\cdot 10=2,4$

$2a)\ \ \sqrt{1,31^2}=1,31\ \ ,\ \ \sqrt{(-1,67)^2}=|-1,67|=1,67\ \ ,\ \ \dfrac{1}{2}\cdot \sqrt{24^2}=\dfrac{24}{2}=12\\\\\\-0,3\sqrt{(-0,9)^2}=-0,3\cdot |-0,9|=-0,3\cdot 0,9=-0,27\\\\\\b)\ \ \sqrt{3^4\cdot 5^2}=3^2\cdot 5=45\\\\\sqrt{(-0,2)^6\cdot (0,2)^4\cdot (-4)^2}=0,2^3\cdot 0,2^2\cdot -4=0,2^5\cdot 4=0,00128\\\\\sqrt{(2-\sqrt3)^2} =|\underbrace {2-\sqrt3}_{>0}|=2-\sqrt3$

$c)\ \ \sqrt{(23-\sqrt7)^2}-\sqrt{(23-\sqrt2)^2}=|\underbrace {23-\sqrt7}_{>0}|-|\underbrace{23-\sqrt2}_{>0}|=\\\\=(23-\sqrt7)-(23-\sqrt2)=\sqrt2-\sqrt7\\\\\\\sqrt{70-14\sqrt{21}}=\sqrt{70-2\cdot 7\sqrt{21}}=\sqrt{(7-\sqrt{21})^2}=|\underbrace{7-\sqrt{21}}_{>0}|=7-\sqrt{21}$

Ответ дал: Аноним

Ответ:

а)

•

$\sqrt{16 \times 81} = \sqrt{1296 } = 36$

•

$\sqrt{ \frac{25}{64} } = \sqrt{ \frac{5}{8} }$

•

$\sqrt{ \frac{9}{49} } \times \sqrt{ \frac{1}{121} } = \sqrt{ \frac{3}{9} } \times \sqrt{ \frac{1}{11} } = \sqrt{ \frac{1}{3} } \times \sqrt{ \frac{1}{11} } = \sqrt{ \frac{1 \times 1}{3 \times 11} } = \sqrt{ \frac{1}{33} } = \frac{1}{ \sqrt{33} }$

б)

•

$\sqrt{50} \times \sqrt{2} = \sqrt{50 \times 2} = \sqrt{100} = 10$

•

$\frac{ \sqrt{343} }{ \sqrt{7} } = \sqrt{49} = 7$

а)

•

$\sqrt{(1.31) {}^{2} } = \sqrt{1.7161} = 1.31$

•

$\sqrt{( - 1.67) {}^{2} } = { \sqrt{2.7889} } = 1.67$

•

$\frac{1}{2} \sqrt{24 {}^{2} } = \frac{1}{2} \sqrt{276} = \frac{1}{2} \times 24 = \frac{1 \times 24}{2} = 12$

•

$- 0.3 \times \sqrt{( - 9) {}^{2} } = - 0.3 \times \sqrt{81} = - 0.3 \times 9 = 2.7$

б)

•

$\sqrt{3 {}^{4} \times 5 {}^{2} } = \sqrt{3 {}^{4} \times 25 } = 3 {}^{2} \times 5 = 9 \times 5 = 45$

•

$\sqrt{( - 2) {}^{6} }$

$\sqrt{( - 2) {}^{6} \times (0.2) {}^{4} \times (4) {}^{2} } = 2 {}^{3} \times 0.2 {}^{2} \times 4 = 2 {}^{3} \times ( \frac{1}{5} ) {}^{2} \times 2 {}^{2} = 2 {}^{5} \times \frac{1}{25} = \frac{2 {}^{5} }{25} = \frac{32}{25} = 1.28$

•

$\sqrt{(2 - \sqrt{3}) {}^{2} } = \sqrt{4 - 3} = \sqrt{1} = 1$

в)

$\sqrt{( \sqrt{} 23 - 7) {}^{2} } - \sqrt{( \sqrt{23} - 2) {}^{2} } = \sqrt{23 - 49} - (\sqrt{23 - 4} ) = 7 - \sqrt{23} - \sqrt{23 + 4} = 7 - \sqrt{23} - \sqrt{23} + 2 = 9 - \sqrt{49} = 9 - 7 = 2$