Question

решыте пожалуйсто поскорее x²=1/9

Answer 1

$\displaystyle\bf\\x^{2} =\frac{1}{9} \\\\x_{1,2} =\pm \ \sqrt{\frac{1}{9} } =\pm \ \sqrt{\Big(\frac{1}{3} \Big)^{2} } =\pm \ \frac{1}{3} \\\\\\Otvet \ : \ x_{1} =-\frac{1}{3} \ \ ; \ \ x_{2} =\frac{1}{3}$