Calculați modulul numărului real x în fiecare din cazurile ex 18 pls dau coroana dar mai tarziu

Răspuns:
[tex]18a) |8 \sqrt{15} + (18 \sqrt{5}) \div ( - 3 \sqrt{3}) ) | \div (2 \sqrt{3} ) = \\ = |8 \sqrt{15} - \frac{18 \sqrt{5} }{3 \sqrt{ 3} } | \div (2 \sqrt{3} ) = \\ = |8 \sqrt{15} - \frac{18 \sqrt{15} }{27} | \div (2 \sqrt{3} ) = \\ = |8 \sqrt{15} - \frac{2 \sqrt{15} }{3} | \div (2 \sqrt{3} ) = \\ = | \frac{24 \sqrt{15} }{3} - \frac{2 \sqrt{15} }{3} | \div (2 \sqrt{3} ) = \\ = | \frac{22 \sqrt{15} }{3} | \div (2 \sqrt{3} ) \\ = \frac{22 \sqrt{15} }{3} \div (2 \sqrt{3} ) = \\ = \frac{22 \sqrt{15} }{3} \times \frac{1}{2 \sqrt{3} } = \frac{22 \sqrt{15} }{6 \sqrt{3} } = \frac{11 \sqrt{5} }{3} [/tex]
[tex] |x| = | \frac{11 \sqrt{5} }{3} | = \frac{11 \sqrt{5} }{3} [/tex]
[tex]b)x = | \frac{40 \sqrt{3} }{ - 2 \sqrt{5} } - 5 \sqrt{15} | \div 3 \sqrt{5} \\ x = | \frac{40 \sqrt{15} }{ - 10} - 5 \sqrt{15} | \div 3 \sqrt{5} \\ x = | - 4 \sqrt{15} - 5 \sqrt{15} | \div 3 \sqrt{5} \\ x = | - 9 \sqrt{15} | \div 3 \sqrt{5} \\ x = 9 \sqrt{15} \div 3 \sqrt{5} \\ x = 3 \sqrt{3} \\ |x| = |3 \sqrt{3} | = 3 \sqrt{3} [/tex]
[tex]c)x = | - {( \sqrt{6} )}^{3} + 8 \sqrt{30} \div ( - \sqrt{5}) | \div ( - 7 \sqrt{3} ) \\ x = | - 6 \sqrt{6} + 8 \times ( - \sqrt{6} ) | \div ( - 7 \sqrt{3} ) \\ x = | - 6 \sqrt{6} - 8 \sqrt{6} | \div ( - 7 \sqrt{3} ) \\ x = | - 14 \sqrt{6} | \div ( - 7 \sqrt{3} ) \\ x = 14 \sqrt{6} \div ( - 7 \sqrt{3} ) \\ x = - 2 \sqrt{2} \\ |x | = | - 2 \sqrt{2} | = 2 \sqrt{2} [/tex]
[tex]d)x = ( - 6 \sqrt{3} ) \div |(3 \sqrt{42}) \div ( - \sqrt{7} ) - ( - { \sqrt{6}) }^{3} | \\ x = ( - 6 \sqrt{3} ) \div | - 3 \sqrt{6} - ( - 6 \sqrt{6}) | \\ x = ( - 6 \sqrt{3} ) \div | - 3 \sqrt{6} + 6 \sqrt{6} | \\ x = ( - 6 \sqrt{3} ) \div |3 \sqrt{6} | \\ x = ( - 6 \sqrt{3} ) \div 3 \sqrt{6} \\ x = \frac{ - 6 \sqrt{3} }{3 \sqrt{6} } \\ x = \frac{ - 6 \sqrt{18} }{18} \\ x = \frac{ - 6 \times 3 \sqrt{2} }{18} \\ x = \frac{ - 18 \sqrt{2} }{18} \\ x = - \sqrt{2} \\ |x| = | - \sqrt{2} | = \sqrt{2} [/tex]