Question

Valendo 50 pontos

Se $\sqrt{9 ^{p+1} } = 3^{ \sqrt{2} }$ e $log_{2} (q-1) = \frac{1}{2}$
Calcule:

$]p^{2} + 2p$ x $q + q^{2}$

Answer 1

$\sqrt{9^{p+1}}=3^{ \sqrt{2} } =\ \textgreater \ \sqrt{9^{p}.9} = 3^{ \sqrt{2} } =\ \textgreater \ p = \sqrt{2} - 1 \\ \\ log_{2}(q-1)= \frac{1}{2} =\ \textgreater \ 2^{ \frac{1}{2} } = (q - 1) =\ \textgreater \ q = \sqrt{2} + 1\\ \\ p^{2}+2p.q + q^{2} \\ \\ ( \sqrt{2} - 1)^{2}+2.( \sqrt{2} -1).( \sqrt{2} +1)+ ( \sqrt{2} +1)^{2} \\ \\ 2 - 2 \sqrt{2} +1+2(2-1)+2+2 \sqrt{2} +1 \\ \\ 2+ 1 + 2 + 2 + 1 = 8$

Espero ter ajudado.