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Calculti:
a)6 supra radica2+7 supra 3radical2-12supra 2radical3+6radical2 supra radical6


Răspuns :

[tex] \frac{6}{ \sqrt{2} } + \frac{7}{ \sqrt{3} } - \frac{12}{2 \sqrt{3} } + \frac{6 \sqrt{2} }{ \sqrt{6} } = \\ = \frac{6 \sqrt{2} }{2} + \frac{7 \sqrt{3} }{3} - \frac{12 \sqrt{3} }{6} + \frac{6 \sqrt{12} }{6} \\ = 3 \sqrt{2} + \frac{7 \sqrt{3} }{3} - 2 \sqrt{3 } + \sqrt{12} \\ ( \sqrt{12 } = 2 \sqrt{3} ) \\ = 3 \sqrt{2} + \frac{7 \sqrt{3} }{3} - 2 \sqrt{3} + 2 \sqrt{3} \\ = 3 \sqrt{2} + \frac{7 \sqrt{3} }{3} \\ = \frac{9 \sqrt{2} + 7 \sqrt{3} }{3} [/tex]