Odpowiedź:
[tex]\huge\boxed{x\in\left<\dfrac{2-\sqrt6}{2};\ \dfrac{2+\sqrt6}{2}\right>}[/tex]
Szczegółowe wyjaśnienie:
[tex](x-1)^2\leq\dfrac{3}{2}\Rightarrow |x-1|\leq\sqrt{\dfrac{3}{2}}\\\\x-1\leq\dfrac{\sqrt3}{\sqrt2}\ \wedge\ x-1\geq-\dfrac{\sqrt3}{\sqrt2}\qquad|+1\\\\x\leq1+\dfrac{\sqrt3}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\ \wedge\ x\geq1-\dfrac{\sqrt3}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}\\\\x\leq1+\dfrac{\sqrt6}{2}\ \wedge\ x\geq1-\dfrac{\sqrt6}{2}\\\\x\leq\dfrac{2+\sqrt6}{2}\ \wedge\ x\geq\dfrac{2-\sqrt6}{2}\Rightarrow x\in\left<\dfrac{2-\sqrt6}{2};\ \dfrac{2+\sqrt6}{2}\right>[/tex]
