Question

rezolvati ex 2,dau coroana urgent​

Answer 1

Răspuns:

$a)x = 2 \sqrt{3} \times (5 \sqrt{3} + 6 \sqrt{3} - 10 \sqrt{3} ) \\ x = 2 \sqrt{3} \times \sqrt{3} \\ x = 2 \times 3 \\ x = 6$

$b)y = ( \frac{3 \sqrt{3} }{2 \sqrt{3} \times \sqrt{3} } - \frac{4 \sqrt{3} }{3 \sqrt{3} \times \sqrt{3} } ) \times \sqrt{24} \times 2 \times 2 \sqrt{2} \\ \\ y = ( \frac{3 \sqrt{3} }{2 \times 3} - \frac{4 \sqrt{3} }{3 \times 3} ) \times 2 \sqrt{6} \times 4 \sqrt{2} \\ y = ( \frac{3 \sqrt{3} }{6} - \frac{4 \sqrt{3} }{9} ) \times 8 \sqrt{12} \\ y =( \frac{9 \sqrt{3} }{18} - \frac{8 \sqrt{3} }{18} ) \times 8 \times 2 \sqrt{3} \\ y = \frac{ \sqrt{3} }{18} \times 16 \sqrt{3} \\ y = \frac{16 \times 3}{18} \\ y = \frac{48}{18} = \frac{8}{3} \\ mg = \sqrt{x \times y} = \sqrt{6 \times \frac{8}{3} } = \sqrt{ \frac{48}{3} } = \frac{ \sqrt{48} }{ \sqrt{3} } = \sqrt{48 } \div \sqrt{3} = \sqrt{16} = 4$