Question

ayuda urgenteeeeeeeeeeeeeeee

Answer 1

$\sqrt{x+6} + \sqrt{x-2} -2\sqrt{x+1} =0$

$\sqrt{x+6}+\sqrt{x-2}=2\sqrt{x+1}$

$(\sqrt{x+6} +\sqrt{x-2})^{2} =(2\sqrt{x+1})^{2}$

$(\sqrt{x+6})^{2} +2\sqrt{x+6}\sqrt{x-2}+(\sqrt{x-2})^{2}=(2\sqrt{x+1})^{2}$

$(\sqrt{x+6}) +2\sqrt{(x+6)(x-2)}+(\sqrt{x-2})=4(x+1)$

$2x+4+2\sqrt{x^{2}+4x-12}=4x+4$

$2\sqrt{x^{2}+4x-12} =2x$

$\sqrt{x^{2}+4x-12} =\frac{2x}{2}$

$(\sqrt{x^{2}+4x-12})^{2} =x^{2}$

$x^{2}+4x-12=x^{2}$

$x^{2} -x^{2} +4x-12=0$

$4x-12=0$

$4x=12$

$x=\frac{12}{4}$

$x=3$

Comprobacion

$\sqrt{x+6} + \sqrt{x-2} -2\sqrt{x+1} =0$

$\sqrt{3+6} + \sqrt{3-2} -2\sqrt{3+1} =0$

$\sqrt{9} + \sqrt{1} -2\sqrt{4} =0$

$3+1-(2)(2)=0$

$3+1-4=0$

$4-4=0$

0=0