Question

Дам 30 баллов срочно пожалуйста ​

Answer 1

1. Вычислите:

$a) 6\sqrt{1,21} - 2(\sqrt{2})^{2} = 6\sqrt{\frac{121}{100} } -2 * 2 = 6 * \frac{11}{10} - 4 = 3 * \frac{11}{5} - 4 = \frac{33}{5} - 4 = \frac{13}{5} = 2\frac{3}{5} = 2,6$

$b) \sqrt{2\frac{1}{4} } - 3\sqrt{5\frac{4}{9} } = \sqrt{\frac{9}{4} }-3\sqrt{\frac{49}{9} } = \frac{2}{3} - 3 * \frac{7}{3} = \frac{2}{3} - 7 = -\frac{11}{2} = -5\frac{1}{2} = - 5,5$

$c) (\sqrt{18} - \sqrt{2})^{2} = (3\sqrt{2} - \sqrt{2} )^{2} = (2\sqrt{2})^{2} = 4 * 2 = 8$

2. Сравните числа:

$a) \sqrt{6} > \sqrt{5}$

$b) \sqrt{1,5} < \sqrt{1\frac{2}{3} }$

3. Упростите:

$a)3\sqrt{2} +\sqrt{50} - \sqrt{18} = 3\sqrt{2} + 5\sqrt{2} - 3\sqrt{2} = 5\sqrt{2}$

$b) (2\sqrt{5} - \sqrt{27})*\sqrt{3}- 2\sqrt{15} = (2\sqrt{5} - 3\sqrt{3} )\sqrt{3} - 2\sqrt{15} = 2\sqrt{15} - 9 - 2\sqrt{15} = -9$

Answer 2

$1)\ \ 6\sqrt{1,21}-2(\sqrt2)^2=6\cdot 1,1-2\cdot 2=6,6-4=2,6\\\\b)\ \ 8\cdot \sqrt{2\dfrac{1}{4}}-3\cdot \sqrt{5\dfrac{4}{9}}=8\cdot \sqrt{\dfrac{9}{4}}-3\cdot \sqrt{\dfrac{49}{9}}=8\cdot \dfrac{3}{2}-3\cdot \dfrac{7}{3}=12-7=5\\\\\\c)\ \ (\sqrt{18}-\sqrt2)^2=18-2\sqrt{36}+2=20-2\cdot 6=20-12=8$

$2)\ \ a)\ \ 6>5\ \ \Rightarrow \ \ \ \sqrt6>\sqrt5\\\\b)\ \ 1,5=1\dfrac{1}{2}=1\dfrac{3}{6}\ <\ 1\dfrac{2}{3}=1\dfrac{4}{6}\ \ \Rightarrow \ \ \ \sqrt{1,5}\, <\, \sqrt{1\dfrac{2}{3}}\\\\\\3)\ \ a)\ \ 3\sqrt2+\sqrt{50}-\sqrt{18}=3\sqrt{2}+5\sqrt{2}-3\sqrt{2}=5\sqrt2\\\\\\b)\ \ (2\sqrt{5}-\sqrt{27})\cdot \sqrt{3}-2\sqrt{15}=(2\sqrt5-3\sqrt3)\cdot \sqrt3-2\sqrt{15}=\\\\=2\sqrt{15}-3\cdot 3-2\sqrt{15}=-9$