Question

vă rog repede vă rog dau coroană​

Answer 1

d)

$\frac{4 - \sqrt{24} }{\sqrt{32} } + \frac{6 \sqrt{6} + 9 }{4 \sqrt{27} } - \frac{6 + 4 \sqrt{6} }{6 \sqrt{12} } - \frac{8 + \sqrt{24} }{3 \sqrt{8} } = \\ \frac{4 - 2 \sqrt{6} }{4 \sqrt{2} } + \frac{3(2 \sqrt{6} + 3 )}{12 \sqrt{3} } - \frac{2(3 + 2 \sqrt{6}) }{12 \sqrt{3} } - \frac{8 + 2 \sqrt{6} }{6 \sqrt{2} } = \\ \frac{2(2 - \sqrt{6} )}{4 \sqrt{2} } + \frac{2 \sqrt{6} + 3 }{4 \sqrt{3} } - \frac{3 + 2 \sqrt{6} }{6 \sqrt{3} } - \frac{2(4 + \sqrt{6} )}{6 \sqrt{2} } = \\ \frac{2 - \sqrt{6} }{2 \sqrt{2} } + \frac{(2 \sqrt{6} + 3) \sqrt{3} }{12} - \frac{(3 + 2 \sqrt{6}) \sqrt{3} }{18} - \frac{4 + \sqrt{6} }{3 \sqrt{2} } = \\ \frac{(2 - \sqrt{6} ) \sqrt{2} }{4} + \frac{2 \sqrt{18} + 3 \sqrt{3} }{12} - \frac{3 \sqrt{3} + 2 \sqrt{18} }{18} - \frac{(4 + \sqrt{6}) \sqrt{2} }{6} = \\ \frac{2 \sqrt{2} - \sqrt{12} }{4} + \frac{6 \sqrt{2} + 3 \sqrt{ 3} }{12} - \frac{3 \sqrt{3} + 6 \sqrt{2} }{18} - \frac{4 \sqrt{2} + \sqrt{12} }{6} = \\ \frac{2 \sqrt{2} - 2 \sqrt{3} }{4} + \frac{3(2 \sqrt{2} + \sqrt{3} ) }{12} - \frac{3( \sqrt{3} + 2 \sqrt{2}) }{18} - \frac{4 \sqrt{2} + 2 \sqrt{3} }{6} = \\ \frac{ \sqrt{2} - \sqrt{ 3} }{2} - \frac{2 \sqrt{2} + \sqrt{3} }{12} - \frac{ \sqrt{3} + 2 \sqrt{2} }{6} = \\ \frac{6( \sqrt{2} - \sqrt{3} ) - (2 \sqrt{2} + \sqrt{3} ) - 2( \sqrt{3} + 2 \sqrt{2}) }{12} = \\ \frac{6 \sqrt{2} - 6 \sqrt{3} - 2 \sqrt{2} - \sqrt{3} - 2 \sqrt{3} - 4 \sqrt{2} }{12} = \\ \frac{0 - 9 \sqrt{3} }{12} = \\ \frac{ - 9 \sqrt{3} }{12} = - \frac{3 \sqrt{3} }{4}$

Answer 2

Răspuns:

Sper ca te am ajutat, dar nu sunt asa sigura pe rezultat. Am încercat. Sper sa fie bun