0
\\\
x< \frac{1}{3}
\\\
8x^2=1-6x+9x^2
\\\
x^2-6x+1=0
\\\
D_1=9-1=8
\\\
x=3-\sqrt{8} =3-2\sqrt{2}
\\\
x \neq 3+\sqrt{8} > \frac{1}{3} " alt=" \sqrt{x} -1+ \sqrt{2x}+6=6
\\\
\sqrt{x} + \sqrt{2x}=1
\\\
x+2x+2 \sqrt{x\cdot2x} =1
\\\
3x+2 \sqrt{2x^2} =1
\\\
2\sqrt{2x^2} =1-3x
\\\
1-3x>0
\\\
x< \frac{1}{3}
\\\
8x^2=1-6x+9x^2
\\\
x^2-6x+1=0
\\\
D_1=9-1=8
\\\
x=3-\sqrt{8} =3-2\sqrt{2}
\\\
x \neq 3+\sqrt{8} > \frac{1}{3} " align="absmiddle" class="latex-formula">
Ответ: 