Question

raiz de 4x+5 -raiz de 3x+1=1 porfa gracias

Answer 1

$\sqrt{4x+5} - \sqrt{3x+1} =1\\ (\sqrt{4x+5} - \sqrt{3x+1})^{2} =1^{2} \\ (\sqrt{4x+5})^{2}-2 \sqrt{(4x+5)(3x+1)}+ ( \sqrt{3x+1})^{2} =1\\4x+5-2 \sqrt{12 x^{2} +19x+5}+3x+1=1\\ 4x+5+3x+1-1=2 \sqrt{12 x^{2}+19x+5}\\7x+5=2 \sqrt{12 x^{2}+19x+5}\\ x^{2} = ( \sqrt{4x+5} - \sqrt{3x+1} =1\\ (\sqrt{4x+5} - \sqrt{3x+1})^{2} =1^{2} \\ (\sqrt{4x+5})^{2}-2 \sqrt{(4x+5)(3x+1)}+ ( \sqrt{3x+1})^{2} =1\\4x+5-2 \sqrt{12 x^{2} +19x+5}+3x+1=1\\ 4x+5+3x+1-1=2 \sqrt{12 x^{2}+19x+5}$
$( 7x+5)^{2}= (2 \sqrt{12 x^{2}+19x+5})^{2} \\49 x^{2} +70x+25=4(12 x^{2}+19x+5)\\49 x^{2} +70x+25=48x^{2}+76x+20\\49 x^{2} +70x+25-48x^{2}-76x-20=0\\ x^{2} -6x+5=0\\(x-5)(x-1)=0$

x=5
x=1

espero te sirva :)

Answer 2

$\sqrt{4x+5} - \sqrt{3x+1} = 1 \\ \\ \sqrt{4x+5} =1+ \sqrt{3x+1}$

Elevando todo al cuadrado y recuciendo términos semejantes 
$4x+ 5 =1+2 \sqrt{3x+1)} + 3x+1 \\ \\ x+3=2 \sqrt{3x+1}$

Nuevamente al cuadrado
$x^2+6x+9= 4(3x+1) \\ \\ x^2+6x+9=12x+4 \\ \\ x^2-6x+5=0$

Factorizando
(x - 5)(x - 1) = 0
       x - 5 = 0
                                x1 = 5
       x - 1 = 0
                                x2 = 1
                                                       S = {1, 5}