Question

me ajudem e uma questao de matematica

16a9ecceb474d6fbfdf59c8cefa2a0db.docx

Answer 1

Oi. 
$\frac{ \sqrt{2} -4}{ \sqrt{3} } (5- \frac{1}{3} ( \sqrt{2}- \sqrt{3})) \\ \\ \frac{ \sqrt{2} -4}{ \sqrt{3} } (5- \frac{ \sqrt{2} }{3} + \frac{ \sqrt{3} }{ 3} ) \\ \\ (\frac{ \sqrt{2} -4}{ \sqrt{3} } )( \frac{15- \sqrt{2} + \sqrt{3} }{3} ) \\ \\ \frac{15 \sqrt{2}- \sqrt{4}+ \sqrt{6}-60+4 \sqrt{2}-4 \sqrt{3} }{3 \sqrt{3} } \\ \\ \frac{19 \sqrt{2}- 62 +\sqrt{6}-4 \sqrt{3} }{3 \sqrt{3} }$
Racionalizando:
$\frac{19 \sqrt{2}- 62 +\sqrt{6}-4 \sqrt{3} }{3 \sqrt{3} } . \frac{3 \sqrt{3} }{3 \sqrt{3}} = \frac{57 \sqrt{6}-186 \sqrt{3}+3 \sqrt{18}-12 \sqrt{9} }{9 \sqrt{9} } = \\ \\ = \frac{57 \sqrt{6}-186 \sqrt{3}+3*3 \sqrt{2}-12*3 }{27} = \frac{19 \sqrt{6} }{9} - \frac{62 \sqrt{3} }{9} + \frac{ \sqrt{2} }{3} - \frac{4}{3} = -7,62$