Question

va rog frumos sa ma ajutati. Calculand 2√24×(3 supra√2-2 supra√3)+√128,se obtine... multumesc frumos

Answer 1

$2 \sqrt{24} *( \frac{{^ \sqrt{2}}^) ~3}{ \sqrt{2}}- \frac{{^ \sqrt{3}}^)~ 2}{\sqrt{3}})+ \sqrt{128} =\\\\ =2 \sqrt{6*4} *( \frac{^3^)3 \sqrt{2} }{2}- \frac{^2^)2 \sqrt{3} }{3} )+ \sqrt{64*2} \\\\ =2*2 \sqrt{6} *( \frac{3*3 \sqrt{2}-2*2 \sqrt{3} }{6} )*8 \sqrt{2} \\\\ =4 \sqrt{6} *( \frac{9 \sqrt{2} -4 \sqrt{3} }{6} )*8 \sqrt{2}$
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