# x+\frac{1}{x}=4. Given: x=2+\sqrt{3}

$x+\frac{1}{x}=4$.
Given: $x=2+\sqrt{3}$
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Concept used:
The concept by which a fraction is rewritten so that the denominator contains only rational numbers. A variety of techniques for rationalizing the denominator are as follows:
Square roots:
$\left(a>0,b>0,c>0\right)$
$\frac{a}{b+\sqrt{c}}=\frac{a}{b+\sqrt{c}}×\frac{b-\sqrt{c}}{b-\sqrt{c}}$
$=\frac{ab-a\sqrt{c}}{{b}^{2}-c}$
Calculation:
$=2+\sqrt{3}+\frac{2-\sqrt{3}}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=2+\sqrt{3}+\frac{2-\sqrt{3}}{1}$
$=4$
Conclusion:
The required result is proved.