Question

Considerando:
$C=\sqrt[3]{\frac{32^{3}+64^{2} }{4^{6}+8^{3} } }$
$M=\sqrt{2018}+1$
$P=(\frac{2018.2019^{2}-x^{2} 2018^{2}.2019}{2018^{2}-2019^{2} } ).(\frac{2018.2019}{2018+2019}) ^{-1}$
$A=M.(\sqrt{2018}-1$

O resultado da expressão $(A^{2}+C.P.A+1)^{\frac{1}{4} }$ será igual a

Answer 1

Resposta:

√2016.

Explicação passo a passo:

Temos que

$C = \sqrt[3]{\frac{2^{15}+2^{12}}{2^{12}+2^{9}}} = \sqrt[3]{\frac{2^{12}(2^{3}+1)}{2^{9}(2^{3}+1)}} = \sqrt[3]{2^{3}}=2\\\\P = 2018.2019.\frac{2019-2018}{(2018+2019).(2018-2019)}.\frac{2018+2019} {2018.2019} = -1\\\\\A =(\sqrt{2018}+1).(\sqrt{2918}-1) = 2018-1=2017,\ logo:\\\\\ (A^{2}+C.P.A+1)^\frac{1}{4}=(2017^{2}+2.(-1).2017+1)^\frac{1}{4}=((2017-1)^{2})^\frac{1}{4}=\sqrt{2016}$