Question

sendo x =$\frac{ {10}^{3} \sqrt{2. {10}^{2} } }{ {10}^{4} \sqrt{4} }$
e y =$\frac{ {(0.25)}^{2} }{{\binom{1}{2}^{4} } \times \sqrt{2}^{3} }$a forma mais simplificada de x + y é:

A) $\frac{ \sqrt{2} }{2}$

B) √2

C) √2 + 2

D) 10√2​

Answer 1

Resposta:

A resposta não bate com nenhuma das alternativas fornecidas.

Explicação passo-a-passo:

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$x+y=\dfrac{10^3 \sqrt{2\;.\;10^2}}{10^4 \sqrt{4}}+\dfrac{0{,}25^2}{\left(\dfrac{1}{2} \\\right)^4\;.\;(\sqrt{2})^3}\\\\\\x+y=\dfrac{10^3}{10^4}\;.\;\dfrac{\sqrt{2\;.\;10^2}}{\sqrt{4}}+\dfrac{\left(\dfrac{1}{4} \right)^2}{\left(\dfrac{1}{2} \right)^4\;.\;(\sqrt{2})^2\;.\;\sqrt{2}}}\\\\\\x+y=\dfrac{1}{10}\;.\;\dfrac{10 \sqrt{2}}{2}+\dfrac{\dfrac{1}{16}}{\dfrac{1}{16} \;.\;2\sqrt{2}}\\\\\\x+y=\dfrac{\sqrt{2}}{2}+\dfrac{1}{2\sqrt{2}}$

$x+y=\dfrac{1}{2}\;.\;\left(\sqrt{2}+\dfrac{1}{\sqrt{2}} \right)\\\\\\x+y=\dfrac{1}{2}\;.\;\left(\sqrt{2}+\dfrac{1}{\sqrt{2}}\;.\;\dfrac{\sqrt{2}}{\sqrt{2}} \right)\\\\\\x+y=\dfrac{1}{2}\;.\;\left(\sqrt{2}+\dfrac{\sqrt{2}}{2} \right)\\\\\\x+y=\dfrac{1}{2}\;.\;\left(\dfrac{2\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2} \right)\\\\\\x+y=\dfrac{1}{2}\;.\;\dfrac{3\sqrt{2}}{2}\\\\\\\boxed{x+y=\dfrac{3\sqrt{2}}{4}} \quad \rightarrow \quad \mathbf{nenhuma\;das\;alternativas}$