Question

Sa se rationalizeze numitorii :
a) 3/3√6 , 12/∛4 , 4/3√2 , 5/∛-25 , 32/3√8
b) 12/√6-√2 , 12/3√2-4 , 15/2√6+3√2 , 6√6/3√6-4√3
c) 6/∛24-∛18 , 18/∛15+∛3 , 1/∛4-∛2+1 .
Ajutati-ma , va rog !!

Answer 1

$a) \frac{3}{3 \sqrt{6} } = \frac{1}{ \sqrt{6} } = \frac{1 \sqrt{6} }{ \sqrt{6} \sqrt{6} } = \frac{ \sqrt{6} }{6}$

$\frac{12}{ \sqrt[3]{4} } = \frac{12}{ \sqrt[3]{ {2}^{2} } } = \frac{12 \sqrt[3]{2} }{ \sqrt[3]{ {2}^{2} \sqrt[3]{2} } } = \frac{12 \sqrt[3]{2} }{ \sqrt[3]{ {2}^{2} \times 2 } } = \frac{12 \sqrt[3]{2} }{ \sqrt[3]{ {2}^{3} } } = \frac{12 \sqrt[3]{2} }{2} = 6 \sqrt[3]{2}$

$\frac{4}{3 \sqrt{2} } = \frac{4 \sqrt{2} }{3 \sqrt{2} \sqrt{2} } = \frac{4 \sqrt{2} }{3 \times 2} = \frac{4 \sqrt{2} }{6} = \frac{2 \sqrt{2} }{3}$

$\frac{5}{ \sqrt[3]{ - 25} } = \frac{5}{ - \sqrt[3]{25} } = - \frac{5}{ \sqrt[3]{25} } = - \frac{5}{ \sqrt[3]{ {5}^{2} } } = - \frac{5 \sqrt[3]{5} }{ \sqrt[3]{ {5}^{2} \sqrt[3]{5} } } = - \frac{5 \sqrt[3]{5} }{ \sqrt[3]{ {5}^{2} \times 5 } } = - \frac{5 \sqrt[3]{5} }{ \sqrt[3]{ {5}^{3} } } = - \frac{5 \sqrt[3]{5} }{5} = - \sqrt[3]{5}$

$\frac{32}{3 \sqrt{8} } = \frac{32}{6 \sqrt{2} } = \frac{16}{3 \sqrt{2} } = \frac{16 \sqrt{2} }{3 \sqrt{2} \sqrt{2} } = \frac{16 \sqrt{2} }{3 \times 2} = \frac{16 \sqrt{2} }{6} = \frac{8 \sqrt{2} }{3}$

$b) \frac{12}{ \sqrt{6} - \sqrt{2} } = \frac{12( \sqrt{6} + \sqrt{2}) }{( \sqrt{6} - \sqrt{2}) \times ( \sqrt{6} + \sqrt{2}) } = \frac{12( \sqrt{6} + \sqrt{2}) }{6 - 2} = \frac{12( \sqrt{6} + \sqrt{2}) }{4} = 3( \sqrt{6} + \sqrt{2}) = 3 \sqrt{6} + 3 \sqrt{2}$

$\frac{12}{3 \sqrt{2} - 4 } = \frac{12(3 \sqrt{2} + 4) }{(3 \sqrt{2} - 4) \times (3 \sqrt{2} + 4) } = \frac{12(3 \sqrt{2} + 4) }{9 \times 2 - 16} = \frac{12(3 \sqrt{2} + 4) }{18 - 16} = \frac{12(3 \sqrt{2} + 4) }{2} = 6(3 \sqrt{2} + 4) = 18 \sqrt{2} + 24$

$\frac{15}{2 \sqrt{6} + 3 \sqrt{2} } = \frac{15(2 \sqrt{6} - 3 \sqrt{2}) }{(2 \sqrt{6} + 3 \sqrt{2}) \times (2 \sqrt{6} - 3 \sqrt{2}) } = \frac{15(2 \sqrt{6} - 3 \sqrt{2}) }{4 \times 6 - 9 \times 2} = \frac{15(2 \sqrt{6} - 3 \sqrt{2}) }{24 - 18} = \frac{15(2 \sqrt{6} - 3 \sqrt{2}) }{6} = \frac{5(2 \sqrt{6} - 3 \sqrt{2}) }{2} = \frac{10 \sqrt{6} - 15 \sqrt{2} }{2}$

$\frac{6 \sqrt{6} }{3 \sqrt{6} - 4 \sqrt{3} } = \frac{6 \sqrt{6} \times (3 \sqrt{6} + 4 \sqrt{3}) }{(3 \sqrt{6} - 4 \sqrt{3}) \times (3 \sqrt{6} + 4 \sqrt{3}) } = \frac{6 \sqrt{6} \times (3 \sqrt{6} + 4 \sqrt{3}) }{9 \times 6 - 16 \times 3} = \frac{6 \sqrt{6} \times (3 \sqrt{6} + 4 \sqrt{3}) }{54 - 48} = \frac{6 \sqrt{6} \times (3 \sqrt{6} + 4 \sqrt{3}) }{6} = \sqrt{6} \times (3 \sqrt{6} + 4 \sqrt{3}) = 18 + 4 \sqrt{18} = 18 + 12 \sqrt{2}$

$c) \frac{6}{ \sqrt[3]{24} - \sqrt[3]{18} } = \frac{6}{2 \sqrt[3]{3} - \sqrt[3]{18} } = \frac{6(4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324}) }{(2 \sqrt[3]{3} - \sqrt[3]{18}) \times (4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324}) } = \frac{6(4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324}) }{8 \times 3 - 18} = \frac{6(4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324}) }{24 - 18} = \frac{6(4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324}) }{6} = 4 \sqrt[3]{9} + 2 \sqrt[3]{54} + \sqrt[3]{324} = 4 \sqrt[3]{9} + 6 \sqrt[3]{2} + 3 \sqrt[3]{12}$

$\frac{18}{ \sqrt[3]{15} + \sqrt[3]{3} } = \frac{18( \sqrt[3]{225} - \sqrt[3]{45} + \sqrt[3]{9}) }{( \sqrt[3]{15} + \sqrt[3]{3}) \times ( \sqrt[3]{225} - \sqrt[3]{45} + \sqrt[3]{9}) } = \frac{18( \sqrt[3]{225} - \sqrt[3]{45} + \sqrt[3]{9}) }{15 + 3} = \frac{18( \sqrt[3]{225} - \sqrt[3]{45} + \sqrt[3]{9}) }{18} = \sqrt[3]{225} - \sqrt[3]{45} + \sqrt[3]{9}$