Question

SE stie ca x+1supra x =4 ,unde x€R. Calculati x la puterea a 2 a +1supra x la puterea a 2a
B)x la puterea a 3a +1supra x la puterea a 3a
C) X la puterea a 4 a +1 supra x la puterea a4 a
Va rog ajutati-ma!

Answer 1

     
$x+ \frac{1}{x} = 4 \\ \\ \frac{ x^{2} +1}{x} =4 \\ \\ x^{2} +1=4x \\ x^{2} -4x+1=0 \\ \\ x_{12}= \frac{4 \pm \sqrt{16-4} }{2} =\frac{4 \pm \sqrt{12} }{2}=\frac{4 \pm 2 \sqrt{3} }{2}=2 \pm \sqrt{3} \\ x_1=\boxed{2 + \sqrt{3}} \\ x_2=\boxed{2 - \sqrt{3}}$


$\text{Calculam pentru ambele solutii:} \\ a_1) \;\;\; x^{2} + \frac{1}{ x^{2}} = \\=(2+\sqrt{3})^2+ \frac{1}{(2+\sqrt{3})^2} = 4+4\sqrt{3}+3 +\frac{1}{4+4\sqrt{3}+3}=7+4\sqrt{3} +\frac{1}{7+4\sqrt{3}}= \\ \\ =7+4\sqrt{3} +\frac{1(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}=7+4\sqrt{3} +\frac{7-4\sqrt{3}}{49-48}= \\ \\ 7+4\sqrt{3} +\frac{7-4\sqrt{3}}{1}=7+4\sqrt{3}+7-4\sqrt{3}=\boxed{14}$


$a_2) \;\;\; x^{2} + \frac{1}{ x^{2}} = \\=(2-\sqrt{3})^2+ \frac{1}{(2-\sqrt{3})^2} = 4-4\sqrt{3}+3 +\frac{1}{4-4\sqrt{3}+3}=7-4\sqrt{3} +\frac{1}{7-4\sqrt{3}}= \\ \\ =7-4\sqrt{3} +\frac{1(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})}=7-4\sqrt{3} +\frac{7+4\sqrt{3}}{49-48}= \\ \\ 7-4\sqrt{3} +\frac{7+4\sqrt{3}}{1}=7-4\sqrt{3}+7+4\sqrt{3}=\boxed{14} \\ =\ \textgreater \ \;\;\text{La punctul a) rezultatul este acelasi indiferent de solutia ecuatiei.}$


La punctele b) si c) te las pe tine sa faci inlocuirile.
E simplu doar ca e mult de calcul.