Question

Alguien lo puede resolverrrrrrrr​

Answer 1

Respuesta:

$\bold{ T = 0}$

Explicación paso a paso:

Identidad de legendre:

$\bold{ (a+b)^2 - (a-b)^2 = 4ab }$

Entonces:

$\bold{ {(a + b + 5)}^{2} + {(a - b + 1)}^{2} = 4(a + 3)(b + 2)}$

$\bold{ {(a + b + 5)}^{2} + {(a - b + 1)}^{2} ={((a + 3) + (b + 2))}^{2} - {((a + 3) - (b + 2))}^{2} }$

$\bold{ {(a + b + 5)}^{2} + {(a - b + 1)}^{2} ={(a + 3+b + 2)}^{2} - {(a + 3- b - 2)}^{2} }$

$\bold{ {(a + b + 5)}^{2} + {(a - b + 1)}^{2} ={(a + b+ 5)}^{2} - {(a - b + 1)}^{2} }$

$\bold{ \cancel{{(a + b + 5)}^{2}} + {(a - b + 1)}^{2} = \cancel{{(a + b+ 5)}^{2}} - {(a - b + 1)}^{2} }$

$\bold{{(a - b + 1)}^{2} = - {(a - b + 1)}^{2} }$

$\bold{{(a - b + 1)}^{2} + {(a - b + 1)}^{2} } = 0$

$\bold{2{(a - b + 1)}^{2} = 0}$

$\bold{{(a - b + 1)}^{2} = 0}$

$\bold{{(a - b + 1)}= 0}$

$\bold{{a + 1}= b}$

$\texttt{CALCULA:}$

$\bold{ T = { ({(b - a)}^{3} - {( \frac{a - 2}{b - 3} )}^{3} ) }^{2012} }$

$\texttt{SUSTITUIR} : \: \: \: \: \bold{b = a+1}$

$\bold{ T = { ({(a + 1- a)}^{3} - {( \frac{a - 2}{a + 1 - 3} )}^{3} ) }^{2012} }$

$\bold{ T = { ({(1)}^{3} - {( \frac{a - 2}{a - 2} )}^{3} ) }^{2012} }$

$\bold{ T = { (1 - {(1)}^{3} )}^{2012} }$

$\bold{ T = { (1 - 1 )}^{2012} }$

$\bold{ T = { (0 )}^{2012} }$

$\bold{ T = 0}$