Question

por fa con procedimiento​

Answer 1

Respuesta:

$\bold{W = 20}$

Explicación paso a paso:

$\bold{Si:} \: \boxed{ \bold{x+\frac{1}{x} = 3}}$

$\bold{Recordar: \: (a + b)^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)}$

$\bold{ {(x + \frac{1}{x}) }^{3} = {3}^{3} }$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} + 3(x \cdot \frac{1}{x} )(x + \frac{1}{x}) = 27}$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} + 3( \frac{x}{x} )( \underline{ \red{x + \frac{1}{x}}}) = 27}$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} + 3(1)( \underline{ \red{3}}) = 27}$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} + 9 = 27}$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} = 27 - 9}$

$\bold{{x}^{3} + {( \frac{1}{x} )}^{3} = 18}$

$\bold{Hallar \: \: el \: \: valor \: \: de:}$

$\bold{W = [ {x}^{x} + {( \frac{1}{x} )}^{ ( \frac{1}{x} ) } ][ {x}^{( \frac{1}{x} )} + {( \frac{1}{x} )}^{x} ]}$

$\bold{W = {x}^{x} \cdot {x}^{( \frac{1}{x} )} + {x}^{x} \cdot {( \frac{1}{x} )}^{x} + {( \frac{1}{x} )}^{ ( \frac{1}{x} ) } \cdot {x}^{( \frac{1}{x} )} + {( \frac{1}{x} )}^{( \frac{1}{x} )} {( \frac{1}{x} )}^{x} }$

$\bold{W = {x}^{x + \frac{1}{x} } + {x}^{x} \cdot {x}^{ - x} + {x}^{ - ( \frac{1}{x} ) } \cdot {x}^{( \frac{1}{x} )} + {( \frac{1}{x} )}^{x + ( \frac{1}{x} ) }}$

$\bold{W = {x}^{3 } + {x}^{x - x}+ {x}^{ - ( \frac{1}{x} ) + \frac{1}{x} } + {( \frac{1}{x} )}^{3}}$

$\bold{W = {x}^{3 } + {x}^{0}+ {x}^{ 0} + {( \frac{1}{x} )}^{3}}$

$\bold{W = {x}^{3 } + 1 + 1 + {( \frac{1}{x} )}^{3}}$

$\bold{W = {x}^{3 } + {( \frac{1}{x} )}^{3} + 2}$

$\bold{Recordar \: \: que: } \: \bold{{x}^{3} + {( \frac{1}{x} )}^{3} = 18}$

$\bold{W = 18 + 2}$

$\bold{W = 20}$