Question

Si: A* =$A^{2}$ + A ; $A^{0}$ =$A^{2}$ +A +1
Además: A* = $\frac{156}{A^{0} }$
Determinar uno de los valores de A:
a) 1
b)2
c)3
d)4
e)5
ayuda pls con procedimiento, doy coronita

Answer 1

Explicación paso a paso:

Datos:

A*= A²+A

A⁰= A²+A+1

reemplazamos los datos en:

$A {}^{*} = \frac{156}{A {}^{0} } \\ \\ A {}^{2} + A = \frac{156}{A {}^{2} + A + 1} \\ \\ (A {}^{2} + A + 1)(A {}^{2} + A) = 156 \\ \\ (A {}^{2} ) (A{}^{2} ) + (A {}^{2}) ( A) + (A {}^{2}) ( A) + (A) ( A) + (1) ( A {}^{2} ) + (1)( A) = 156 \\ \\ A {}^{4} + 2 A {}^{3} + 2A {}^{2} + A = 156 \\ \\ A {}^{4} + 2 A {}^{3} + 2A {}^{2} + 52 A - 51A - 156 = 0 \\ \\ A {}^{4} - 3 A {}^{3} + 5 A {}^{3} - 15A {}^{2} + 17A {}^{2} + 52 A - 51A - 156 = 0 \\ \\( A {}^{4} - 3 A {}^{3}) + (5 A {}^{3} - 15A {}^{2} )+ (17A {}^{2} - 51 A) + ( 52A - 156) = 0 \\ \\A {}^{3} ( A - 3) + 5 A {}^{2} ( A - 3 )+17 A( A - 3 ) +52 ( A - 3) = 0 \\ \\ (A - 3)(A {}^{3} + 5A {}^{2} + 17A + 52) = 0 \\ \\ (A - 3)(A {}^{3} + 4A {}^{2} + A {}^{2} + 4A + 13A + 52) = 0 \\ \\ (A - 3) [ (A {}^{3} +4A {}^{2}) + ( A {}^{2} +4 A) + (13A + 52)] = 0 \\ \\ (A - 3)[A {}^{2} (A +4 ) +A ( A +4 ) + 13(A + 4)] = 0 \\ \\ (A - 3)(A + 4)(A {}^{2} + A + 13 = 0) = 0 \\ \\ A _1 - 3 = 0 \\ A _1 = 3 \\ \\ A _2 + 4 = 0 \\ A _2 = - 4 \\ \\ A {}^{2} + A + 13 = 0 \: \: \: A \not ∈ \: \mathbb{R}$

uno de los valores de A= 3

Espero que sea de tu ayuda

Saludos :)