Question

$frac{3+x}{3x}= sqrt{ frac{1}{9}+ frac{1}{x} sqrt{ frac{4}{9} + frac{2}{x^2} } }$

Answer 1

x≠0

$frac{3+x}{3x}= sqrt{ frac{1}{9}+ frac{1}{x} sqrt{ frac{4}{9} + frac{2}{x^2} } }\ \ (frac{3+x}{3x})^2=frac{1}{9}+ frac{1}{x} sqrt{ frac{4}{9} + frac{2}{x^2} }\ \ frac{x^2+6x+9-x^2}{9x^2} =frac{1}{x} sqrt{ frac{4}{9} + frac{2}{x^2} }\ \ frac{6x+9}{9x^2} =frac{1}{x} sqrt{ frac{4}{9} + frac{2}{x^2} }\ \ frac{6x^2+9x}{9x^2} =sqrt{ frac{4}{9} + frac{2}{x^2} }\ \frac{2x+3}{3x} =sqrt{ frac{4}{9} + frac{2}{x^2} }\ \frac{4x^2+12x+9}{9x^2} =frac{4x^2+18}{9x^2} \ \12x=9$

$x=frac{3}{4} =0.75$