Question

si:
$x = 2 + \sqrt{3}$
calcule el valor de
$R = {x}^{2} + \frac{1}{ {x}^{2} }$
​

Answer 1

Respuesta:

14

Explicación paso a paso:

Se sabe que:

$x = 2 + \sqrt{3}$

Entonces podremos hallar el valor de x + 1/x

$x + \frac{1}{x} = (2 + \sqrt{3} ) + \frac{1}{(2 + \sqrt{3}) }$

$x + \frac{1}{x} = (2 + \sqrt{3} ) + \frac{1(2 - \sqrt{3} )}{(2 + \sqrt{3})(2 - \sqrt{3}) }$

$x + \frac{1}{x} = 2 + \sqrt{3}+ \frac{2 - \sqrt{3}}{ {2}^{2} - { \sqrt{3} }^{2} }$

$x + \frac{1}{x} = 2 + \sqrt{3}+ \frac{2 - \sqrt{3}}{4 - 3}$

$x + \frac{1}{x} = 2 + \sqrt{3}+ \frac{2 - \sqrt{3}}{1}$

$x + \frac{1}{x} = 2 + \sqrt{3}+2 - \sqrt{3}$

$x + \frac{1}{x} = 2+2 + \sqrt{3} - \sqrt{3}$

$x + \frac{1}{x} = 4$

Elevarlo al cuadrado, tener en cuenta que:

(a+b)² = a²+2ab+b²

$(x+\frac{1}{x})^{2}= (4)^{2}$

${(x)}^{2} + 2(x)( \frac{1}{x}) + {( \frac{1}{x} )}^{2} =16$

${x}^{2} + 2( \frac{x}{x}) + \frac{1}{ {x}^{2} } =16$

${x}^{2} + 2(1) + \frac{1}{ {x}^{2} } =16$

${x}^{2} + 2+ \frac{1}{ {x}^{2} } =16$

${x}^{2} + \frac{1}{ {x}^{2} } =16 - 2$

${x}^{2} + \frac{1}{ {x}^{2} } =14$