Question

$( \frac{4}{ \sqrt{12} } - \frac{6 }{ \sqrt{27} } + \frac{8}{ \sqrt{32} } ): \sqrt{2}$

Answer 1

(4/√12-6/√27+8/√32):√2=
=(4/2√3-6/3√3+8/4√2):√2=
=(2/√3-2/√3+2/√2):√2=
=2/√2:√2=
=2/√2 ×1/√2=
=2/2=1

Answer 2

$( \frac{4}{ \sqrt{12} } - \frac{6}{ \sqrt{27} } + \frac{8}{ \sqrt{32} } ): \sqrt{2} = \\ ( \frac{4}{2 \sqrt{3} } - \frac{6}{3\sqrt{3} } + \frac{8}{4 \sqrt{2} } ): \sqrt{2} = \\ ( \frac{2}{ \sqrt{3} } - \frac{2}{ \sqrt{3} }+ \frac{2}{ \sqrt{2} }): \sqrt{2}= \\ \frac{2}{ \sqrt{2} } : \sqrt{2} = \frac{2}{ \sqrt{2} } * \frac{1}{ \sqrt{2} } = \frac{2}{2} =1$

/robèrtinni