Question

se $x^{2}$ +1/$x^{2}$ =14, com x>0, então $(x+1/x)^{5}$ é igual a:

Answer 1

$\underbrace{\left(x+\frac{1}{x}\right).\left(x+\frac{1}{x}\right)}_{(x~.~x~+~2~.~x~.~\frac{1}{x}~+~\frac{1}{x}~.~\frac{1}{x})}.\underbrace{\left(x+\frac{1}{x}\right).\left(x+\frac{1}{x}\right)}_{(x~.~x~+~2~.~x~.~\frac{1}{x}~+~\frac{1}{x}~.~\frac{1}{x})}.\underbrace{\left(x+\frac{1}{x}\right)}_{\left(x+\frac{1}{x}\right)}~=~\left(x+\frac{1}{x}\right)^5$

$\left(x^2~+~2\,.\,\frac{x}{x}~+~\frac{1}{x^2}\right)~.~\left(x^2~+~2\,.\,\frac{x}{x}~+~\frac{1}{x^2}\right)~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5\\\\\\\left(x^2+2+\frac{1}{x^2}\right)~.~\left(x^2+2+\frac{1}{x^2}\right)~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5\\\\\\\left(14+2\right)~.~\left(14+2\right)~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5$

$\left(16\right)~.~\left(16\right)~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5\\\\\\16^2~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5\\\\\\256~.~\left(x~+~\frac{1}{x}\right)~=~\left(x~+~\frac{1}{x}\right)^5\\\\\\\boxed{256x+\frac{256}{x}~=~\left(x~+~\frac{1}{x}\right)^5}$