Question

alguém sabe responder isso ?
( com o cálculo)

Answer 1

Olá!



$\dfrac{1}{\sqrt{x+\sqrt{x^2-1}}}+\dfrac{1}{\sqrt{x-\sqrt{x^2-1}}}=\sqrt{2(x^2+1)}\\\\\dfrac{\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}}{\sqrt{(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})}}=\sqrt{2(x^2+1)}\\\\\dfrac{\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}}{\sqrt{x^2-\sqrt{x^2-1}^2}}=\sqrt{2(x^2+1)}\\\\\dfrac{\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}}{\sqrt{1}}=\sqrt{2(x^2+1)}\\\\\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=\sqrt{2(x^2+1)}$


Todo radical duplo pode ser expresso pela soma ou subtração de dois radicais.
$\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+c}{2}}\pm\sqrt{\dfrac{a-c}{2}},\qquad\text{em que: }c=a^2-b$
Aplicando nos radicais duplos da equação:
$\sqrt{x-\sqrt{x^2-1}}=\sqrt{\dfrac{x+1}{2}}-\sqrt{\dfrac{x-1}{2}}\\\\\sqrt{x+\sqrt{x^2-1}}=\sqrt{\dfrac{x+1}{2}}+\sqrt{\dfrac{x-1}{2}}$


Substituindo na equação:
$\sqrt{x-\sqrt{x^2-1}}+\sqrt{x+\sqrt{x^2-1}}=\sqrt{2(x^2+1)}\\\\\left(\sqrt{\dfrac{x+1}{2}}-\sqrt{\dfrac{x-1}{2}}\right)+\left(\sqrt{\dfrac{x+1}{2}}+\sqrt{\dfrac{x-1}{2}}\right)=\sqrt{2(x^2+1)}\\\\2\sqrt{\dfrac{x+1}{2}}=\sqrt{2(x^2+1)}\\\\\left(2\sqrt{\dfrac{x+1}{2}}\right)^2=(\sqrt{2(x^2+1)})^2\\\\4\left(\dfrac{x+1}{2}\right)=2(x^2+1)\\\\\boxed{x=0\quad\text{ou}\quad{x}=1}\\\\\boxed{\boxed{S=\{0,\,\,1\}}}$



Bons estudos!