Matemáticas
bobis
1

Resuelve: (1/√28 + √(4/7)) * 14/5 * √(4/14) ..... Por favor necesito ahorita.

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(2) Respuestas
fani2013

[latex]( \frac{1}{ \sqrt{28} }+ \sqrt{ \frac{4}{7} })* \frac{14}{5}* \sqrt{ \frac{4}{14} } = \\ \\ ( \frac{1}{ \sqrt{7*4} }+ \frac{2}{ \sqrt{7} })* \frac{14}{5}* \frac{2}{ \sqrt{14} }= \\ \\ ( \frac{1}{2 \sqrt{7} }+ \frac{2}{ \sqrt{7} })* \frac{14}{5}* \frac{2}{ \sqrt{14} } = \\ \\ \frac{1+4}{2 \sqrt{7} }* \frac{14}{5}* \frac{2}{ \sqrt{14} }= \\ \\ \frac{5}{2 \sqrt{7} }* \frac{14}{5}* \frac{2}{ \sqrt{14} }= \\ \\ \frac{5*14*2}{2 \sqrt{7} *5* \sqrt{14} }= [/latex] [latex] \frac{14*2}{2 \sqrt{7}* \sqrt{14} }= \\ \\ \frac{28}{2 \sqrt{7*14} }= \\ \\ \frac{28}{2 \sqrt{98} }* \frac{ \sqrt{98} }{ \sqrt{98} }= \frac{28 \sqrt{98} }{2*98}= \frac{14 \sqrt{98} }{98} [/latex] [latex] \frac{7 \sqrt{98} }{49} = \frac{ \sqrt{98} }{7} [/latex] [latex] \frac{ \sqrt{49*2} }{7}= \frac{7 \sqrt{2} }{7}= \sqrt{2} [/latex]

ariannimontoya

Veamos: (1/√28+√4/7) x 14/5 x √4/14 (1/2√7+2/√7) x 14/5 x 2/√14 5/2√7 x 14/5 x 2/√14   (5x14x2) _________ 2√7x5x√14   28 ____ 2√98   14     √98 ____ x ___ √98     √98 14√98 _____    98 7√98 ____   49 7x7√2 _____    49 √2 = respuesta! Buen dia! Atte:Cazador.

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