Question

se n+1/n=12, então n^2+1/n^2, vale?

Answer 1

$n+\frac{1}{n}=12\\\\MMC=n\\\\\frac{n.n+1}{n}=12\\\\n^2+1=12n\\\\n^2-12n+1=0\\\\Bhaskara\\\Delta = (-12)^2-4.1.1\\\Delta=144-4\\\Delta=140\\\\n_1=\frac{12+\sqrt{140}}{2.1}=\frac{12+2\sqrt{35}}{2}=6+\sqrt{35}\\\\n_2=\frac{12-\sqrt{140}}{2.1}=\frac{12-2\sqrt{35}}{2}=6-\sqrt{35}$

Substituindo o n1 (ou n2) na 2ª equação:

$(6+\sqrt{35})^2+\frac{1}{(6+\sqrt{35})^2}\\\\(6^2+6\sqrt{35}+6\sqrt{35}+\sqrt{35}^2)+\frac{1}{(6^2+6\sqrt{35}+6\sqrt{35}+\sqrt{35}^2)}\\\\(36+12\sqrt{35}+35)+\frac{1}{(36+12\sqrt{35}+35)}\\\\(71+12\sqrt{35})+\frac{1}{(71+12\sqrt{35}}\\\\(71+12\sqrt{35})+\frac{1}{(71+12\sqrt{35}}.\frac{71-12\sqrt{35}}{71-12\sqrt{35}}\\\\(71+12\sqrt{35})+\frac{71-12\sqrt{35}}{71^2-12^2.35}\\\\(71+12\sqrt{35})+\frac{71-12\sqrt{35}}{5041-5040}\\\\(71+12\sqrt{35})+\frac{71-12\sqrt{35}}{1}$

$71+12\sqrt{35}+71-12\sqrt{35}\\\\142$

Resp: 142