Question

ayuda (URGENTE POR FAVOR..............)​

Answer 1

Respuesta:

H = 1

Explicación paso a paso:

saber que:

$z.z.z.z.......(k \: \: veces)= z^k$

$\frac{ {z}^{a} }{ {z}^{b} } = {z}^{a - b}$

$\frac{1}{ {z}^{a} } = {z}^{ - a}$

$({z}^{a})( {z}^{b} ) = {z}^{a + b}$

${( {z}^{a} )}^{b} = {z}^{ab}$

En el problema:

$H = ( \frac{ {z}^{(3n + 6)} }{ {z}^{(4n - 2)} } )( \frac{ {z}^{(n + 3)} }{ {z}^{6} } )^{2} ( \frac{1}{ {z}^{(n + 2)} } )$

$H = ({z}^{(3n + 6) - (4n - 2)} )( {z}^{(n + 3) - 6})^{2} ({z}^{ - (n + 2)})$

$H = ({z}^{3n + 6-4n + 2} )( {z}^{n + 3 - 6})^{2} ({z}^{ -n - 2})$

$H = ({z}^{ - n + 8} )( {z}^{n - 3})^{2} ({z}^{ -n - 2})$

$H = ({z}^{ - n + 8} )( {z}^{2(n - 3)}) ({z}^{ -n - 2})$

$H = ({z}^{ - n + 8} )( {z}^{2n - 6}) ({z}^{ -n - 2})$

$H = {z}^{ (- n + 8) + (2n - 6) + (-n - 2)}$

$H = {z}^{ - n + 8+ 2n - 6-n - 2}$

$H = {z}^{ - n + 2n - n+ 8 - 6 - 2}$

$H = {z}^{ - n + n+ 2- 2}$

$H = {z}^{0+0}$

$H = {z}^{0}$

$H =1$