Question

Alguém responde essa questão, Urgente

Answer 1

Resposta:

286.x²

Explicação passo a passo:

$(\sqrt{x} + \frac{1}{x}) ^{13}\\ \\ Termo \ Geral (TG): \binom{13}{p}\sqrt{x}^{13-p}.\frac{1}{x}^p\\ \\ (TG): \binom{13}{p}(x^{\frac{1}{2}})^{13-p}.(x^{-1})^p\\ \\ (TG): \binom{13}{p}x^{\frac{13-p}{2}}.x^{-p}\\ \\ (TG): \binom{13}{p}x^{\frac{13-p}{2} + (-p)}\\ \\ (TG): \binom{13}{p}x^{\frac{13-p-2p}{2}}\\ \\ (TG): \binom{13}{p}x^{\frac{13-3p}{2}}\\ \\ Para \ x^2....\\ \\ \frac{13-3p}{2} = 2\\ 13-3p = 4\\ p = 3$

$Logo...\\ \\ (TG): \binom{13}{p}x^{\frac{13-3p}{2}}\\ \\ (Tx^2): \binom{13}{3}x^{\frac{13-3.3}{2}}\\ \\ (Tx^2): \frac{13.12.11.10!}{(13-3)!.3!}x^{\frac{4}{2}}\\ \\ \boxed{(Tx^2): 286x^{2}}$