0\\
x>0\\\\
2x=(1-x)^2\\
2x=1-2x+x^2\\
1-4x+x^2=0\\
x^2-4x+1=0\\
D=\sqrt{12}\\
x=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\
x=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}" alt="\sqrt{2x}=1-x\\
2x>0\\
x>0\\\\
2x=(1-x)^2\\
2x=1-2x+x^2\\
1-4x+x^2=0\\
x^2-4x+1=0\\
D=\sqrt{12}\\
x=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\
x=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}" align="absmiddle" class="latex-formula">
Ответ
