Question

si: (a+1)²=(√3+2)a
calcular: N= (a²+1)²/ a⁴+1

Answer 1

El valor de N dada (a+1)²=(√3+2)a es:

N = 1

N = 3

Despejar a;

(a+1)²=(√3+2)a

Aplicar binomio cuadrado;

(a+1)² = a² + 2a + 1

Sustituir;

a² + 2a + 1 = √3a + 2a

a² + 2a + 1 - √3a - 2a = 0

Agrupar;

a² -√3a + 1 = 0

Aplicar la resolvente;

$x_{1} = \frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$x_{2} = \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Siendo;

a= 1

b= -√3

c = 1

Sustituir;

$x_{1} = \frac{\sqrt{3}+\sqrt{\sqrt{3}^{2}-4(1)(1) } }{2(1)}$

$x_{1} = \frac{\sqrt{3}+\sqrt{3-4} }{2}$

$x_{1} = \frac{\sqrt{3}+i }{2}$

$x_{2} = \frac{\sqrt{3}-i }{2}$

Entonces;

a = $\frac{\sqrt{3}+i }{2}$

a =  $\frac{\sqrt{3}-i }{2}$

Sustituir los valores de a en N;

• Para a = $\frac{\sqrt{3}+i }{2}$;

N =  $\frac{(a^{2}+1)^{2}  }{a^{4} +1}$

N =$\frac{((\frac{(\sqrt{3}+i }{2})^{2}+1)^{2}  }{(\frac{\sqrt{3}+i }{2})^{4} +1}$

N =$\frac{((\frac{1+\sqrt{3} i }{2}+1)^{2}  }{(\frac{1+\sqrt{3} }{2} +1}$

N = $\frac{((\frac{1+\sqrt{3}i}{2} }{(\frac{1+\sqrt{3} i }{2} +1}$

N = 1

• Para a =  $\frac{\sqrt{3}-i }{2}$

N =  $\frac{(a^{2}+1)^{2}  }{a^{4} +1}$

N =$\frac{((\frac{(\sqrt{3}-i }{2})^{2}+1)^{2}  }{(\frac{\sqrt{3}-i }{2})^{4} +1}$

N =$\frac{((\frac{1-\sqrt{3} i }{2}+1)^{2}  }{(\frac{-1-\sqrt{3}i }{2} +1}$

N = $\frac{((\frac{3-\3sqrt{3}i}{2} }{(\frac{1-\sqrt{3} i }{2} }$

N = 3