Question

Fie numerele a=radical 6•

Answer 1

Uite aici rezolvarea, sper sa intelegi

Answer 2

Răspuns:

$a = \sqrt{6} ( \frac{3}{ \sqrt{2} } + \frac{2}{ \sqrt{3} } ) - \sqrt{27} \\ a = \sqrt{6}( \frac{3 \sqrt{2} }{2} + \frac{2 \sqrt{3} }{3} ) - 3 \sqrt{3} \\ a = \frac{3 \sqrt{12} }{2} + \frac{2 \sqrt{18} }{3} - 3 \sqrt{3} \\ = \frac{6 \sqrt{3} }{2} + \frac{6\sqrt{2} }{3} - 3 \sqrt{3} \\ = 3 \sqrt{3} + 2 \sqrt{2} - 3 \sqrt{3} = 2 \sqrt{2} \\ b = \sqrt{12} + \sqrt{18} - 2 \sqrt{3} \\ b = 2 \sqrt{3} + 3 \sqrt{2} - 2 \sqrt{3} = 3 \sqrt{2} \\ ma = \frac{2 \sqrt{2} + 3 \sqrt{2} }{2} = \frac{5 \sqrt{2} }{2}$