Question

Alguien me puede ayudar porfavor ?

Answer 1

Respuesta:

꧁Holi꧂

$\frac{ {a}^{2} - 6a + 10 }{ {a}^{2} + 8a + 17 } = ( \frac{a - 3}{a + 4} )$

$\frac{ {a}^{2} - 6a + 10}{ {a}^{2} + 8a + 17 } = ( \frac{a - 3}{a + 4} {)}^{2} ,a ≠ - 4$

$\frac{ {a}^{2} - 6a + 10 }{ {a}^{2} + 8a + 17 } = \frac{(a - 3 {)}^{2} }{(a + {4)}^{2} }$

$\frac{ {a}^{2} - 6a + 10 }{ {a}^{2} + 8a + 17} = \frac{ {a}^{2} - 6a + 9}{(a + {4)}^{2} }$

$( {a}^{2} - 6a + 10) \times (a+ 4 {)}^{2} = ( {a}^{2} - 6a + 9) \times ( {a}^{2} + 8a + 17)$

$( {a}^{2} - 6a + 10) \times ( {a}^{2} + 8a + 16 = ( {a}^{2} - 6a + 9) \times ( {a}^{2} + 8a + 17)$

$( {a}^{2} - 6a + 10) \times ( {a}^{2} + {a}^{3} + {17a}^{2} - {6a}^{3} - {48a}^{2} - 102a + {9a}^{2} + 72a + 153$

${a}^{4} + {8a}^{2} + {16a}^{2} - {6a}^{3} - {48a}^{2} - 96a + {10a}^{2} + 80a + 160 = {8}^{4} + {8a}^{3} + {17a}^{2} - {6a}^{3} - {48a}^{2} - 102a + {5a}^{2} + 72a + 153$

{8a}^{3} + {8a}^{2} + {16a}^{2} - {6a}^{3} - {48a}^{2} - 96a + {10a}^{2} + 80a + 160 = {8}^{4} + {8a}^{3} + {17a}^{2} - {6a}^{3} - {48a}^{2} - 102a + {5a}^{2} + 72a + 153

{16a}^{2} + {8a}^{2} + {16a}^{2} - {6a}^{3} - {48a}^{2} - 96a + {10a}^{2} + 80a + 160 = {8}^{4} + {8a}^{3} + {17a}^{2} - {6a}^{3} - {48a}^{2} - 102a + {5a}^{2} + 72a + 153

${16a}^{2} - {48a}^{2} - 96a + {10a}^{2} + 80a + 160 = {17a}^{2} - {48a}^{2} - 102a + {9a}^{2} + 72a + 153$

{16a}^{2} - 96a - 96a + {10a}^{2} + 80a + 160 = {17a}^{2} - {48a}^{2} - 102a + {9a}^{2} + 72a + 153

${26a}^{2} - 96a + 80a + 160 = {17a}^{2} - 102a + {9a}^{2} + 72a + 153$

{26a}^{2} - 16a + 80a + 160 = {17a}^{2} - 102a + {9a}^{2} + 72a + 153

{26a}^{2} - 96a + 80a + 160 = {26a}^{2} - 102a + {9a}^{2} + 72a + 153

{26a}^{2} - 96a + 80a + 160 = {26a}^{2} - 30a + 153

$- 16a + 160 = - 30a + 153$

$- 16a + 30a + 160 = 153$

$- 16a + 30a = 153 - 160$

$14a = 153 - 160$

$14a = - 7$

$a = - \frac{1}{2}$

Rpt: -1/2