Question

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

Answer 1

x≠0

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

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