Question

va rog ... dau coroana

Answer 1

$a)x = \frac{15}{ \sqrt{3} } \\ \\ x = \frac{15 \sqrt{3} }{ \sqrt{3} \sqrt{3}} \\ \\ x = \frac{15 \sqrt{3} }{3} \\ \\ x = 5 \sqrt{3} \\ \\ \\ y = \frac{15}{ \sqrt{5} } \\ \\ y = \frac{15 \sqrt{5} }{ \sqrt{5} \sqrt{5} } \\ \\ y = \frac{15 \sqrt{5} }{5} \\ \\ y = 3 \sqrt{5} \\ \\ \\ = > x > y$

$b)x = 2 \sqrt{8} + 4 \sqrt{50} \\ \\ x = 4 \sqrt{2} + 20 \sqrt{2} \\ \\ x = (4 + 20) \sqrt{2} \\ \\ x = 24 \sqrt{2} \\ \\ \\ y = 6 \sqrt{6} + 3 \sqrt{24} \\ \\ y = 6 \sqrt{6} + 6 \sqrt{6} \\ \\ y = (6 + 6) \sqrt{6} \\ \\ y = 12 \sqrt{6} \\ \\ = > x > y$

$c)x = ( \sqrt{54} + \sqrt{27}) : 3 \\ \\ x = (3 \sqrt{6} + 3 \sqrt{3}) : 3 \\ \\ x = \frac{3 \sqrt{6} + 3 \sqrt{3} }{3} \\ \\ x = \frac{3( \sqrt{6} + \sqrt{3}) }{3} \\ \\ x = \sqrt{6} + \sqrt{3} \\ \\ \\ y = ( \sqrt{150} + \sqrt{125}) : 5 \\ \\ y = (5 \sqrt{6} + 5 \sqrt{5}) : 5 \\ \\ y = \frac{5 \sqrt{6} + 5 \sqrt{5} }{5} \\ \\ y = \frac{5( \sqrt{6} + \sqrt{5}) }{5} \\ \\ y = \sqrt{6} + \sqrt{5} \\ \\ = > x < y$

$( \frac{2}{ \sqrt{45} } - \frac{4}{ \sqrt{125} } + \frac{3}{ \sqrt{20} }) \times \frac{18 \sqrt{5} }{41} = ( \frac{2}{3 \sqrt{5} } - \frac{4}{5 \sqrt{5} } + \frac{3}{2 \sqrt{5} }) \times \frac{18 \sqrt{5} }{41} = ( \frac{2 \sqrt{5} }{3 \times 5} - \frac{4 \sqrt{5} }{5 \times 5} + \frac{3 \sqrt{5} }{2 \times 5}) \times \frac{18 \sqrt{5} }{41} = ( \frac{2 \sqrt{5} }{15} - \frac{4 \sqrt{5} }{25} + \frac{3 \sqrt{5} }{10}) \times \frac{18 \sqrt{5} }{41} = \frac{41 \sqrt{5} }{150} \times \frac{18 \sqrt{5} }{41} = \frac{41 \sqrt{5} \times 18 \sqrt{5} }{150 \times 41} = \frac{738 \sqrt{5} \sqrt{5} }{150 \times 41} = \frac{738 \sqrt{25} }{150 \times 41} = \frac{738 \times 5}{150 \times 41} = \frac{3690}{6150} = \frac{3}{5}$