Question

x²-2x-1≥0 alguém me ajuda ​

Answer 1

$\mathsf{{x}^{2}-2x-1\ge\,0}$

fazendo $\mathsf{f(x)={x}^{2}-2x-1}$

vamos fazer o estudo do sinal

$\mathsf{{x}^{2}-2x-1=0}\\\mathsf{\Delta={(-2)}^{2}-4.1.(-1)=4+4=8}\\\mathsf{x=\dfrac{-(-2)\pm\sqrt{8}}{2.1}=\dfrac{2\pm2\sqrt{2}}{2}}$

$\mathsf{x=1\pm\sqrt{2}}\\\mathsf{x_{1}=1+\sqrt{2}}\\\mathsf{x=1-\sqrt{2}}$

$\mathsf{f(x)>0\to\,x\textless\,1-\sqrt{2}\:\:ou\:x>1+\sqrt{2}}$

$\mathsf{f(x)\textless\,0\to\,1-\sqrt{2}\textless\,x\textless\,1+\sqrt{2}}$

A inequação pede para que valor de x

$\mathsf{f(x)\ge0}$

portanto

$\mathsf{f(x)\ge0\to\,x\le\,1-\sqrt{2}\:\:ou\:x\ge1+\sqrt{2}}$