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Question Number 173188 by dragan91 last updated on 08/Jul/22

Answered by mr W last updated on 08/Jul/22

$$t=\sqrt{x^2+x+1}>0$$$$\frac{2t}{\sqrt{t^2+1}}+\frac{2(t^2+1)}{t}=4+\sqrt{2}$$$$\frac{1}{\sqrt{1+\frac{1}{t}}}+t+\frac{1}{t}=2+\frac{1}{\sqrt{2}}$$$$f\left(t\right)=\frac{1}{\sqrt{1+\frac{1}{t}}}+t+\frac{1}{t}⩾2+\frac{1}{\sqrt{2}}$$$$‘‘{=}^{″}holdsatt=1$$$$\Rightarrow solutionist=\sqrt{x^2+x+1}=1$$$$x(x+1)=0$$$$\Rightarrow x=0or-1✓$$

Commented by dragan91 last updated on 08/Jul/22

$$\mathrm{nice}$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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