Question

Ajuda matemática valor numérico da expressão , tem que ter a conta

Answer 1

vamos lá...

$\sqrt{ \frac{3}{4}- \frac{1}{12} } + \sqrt{2.( \frac{1}{12}) } - \frac{3}{2} \sqrt{1-4.( \frac{1}{12}) } = \\ \\ \sqrt{ \frac{9-1}{12} } + \sqrt{\not2. \frac{1}{\not12_6} } - \frac{3}{2} \sqrt{1-\not4. \frac{1}{\not12_3} } = \\ \\ \sqrt{ \frac{8}{12} } + \sqrt{ \frac{1}{6} } - \frac{3}{2} \sqrt{1- \frac{1}{3} } = \\ \\ \frac{ \sqrt{8} }{ \sqrt{12} } + \frac{1}{ \sqrt{6} } - \frac{3}{2} \sqrt{ \frac{3-1}{3} } =$ 

$\frac{ \sqrt{2^2.2} }{ \sqrt{2^2.3} } + \frac{1}{ \sqrt{6} } - \frac{3}{2} \sqrt{ \frac{2}{3} } = \\ \\ \frac{2 \sqrt{2} }{2 \sqrt{3} } + \frac{1}{ \sqrt{6} } - \frac{3 \sqrt{2} }{2 \sqrt{3} } \\ \\ ( \frac{2 \sqrt{2}-3 \sqrt{2} }{2 \sqrt{3} } )+ \frac{1}{ \sqrt{6} } = \\ \\ - \frac{ \sqrt{2} }{2 \sqrt{3} } + \frac{1}{ \sqrt{6} } = \\ \\ racionalizar \\ \\ - \frac{ \sqrt{2}. \sqrt{3} }{2. \sqrt{3}. \sqrt{3} } =- \frac{ \sqrt{6} }{2 \sqrt{9} } =- \frac{ \sqrt{6} }{2.3} =- \frac{ \sqrt{6} }{6}$ 

$\frac{ \sqrt{6} }{ \sqrt{6}. \sqrt{6} } = \frac{ \sqrt{6} }{ \sqrt{36} } = \frac{ \sqrt{6} }{6} \\ \\ logo \\ \\ - \frac{ \sqrt{6} }{6} + \frac{ \sqrt{6} }{6} =\fbox{ 0 } \\ \\ Letra~~D$