Question

Помогите решить
$x \frac{1}{2}$
, если:
$\frac{4}{9} \leqslant x \leqslant 1 \frac{11}{25}$
​

Answer 1

$\frac{4}{9}\leq x\leq 1\frac{11}{25}\\\\\frac{4}{9} \leq x\leq \frac{36}{25}\\\\(\frac{4}{9})^{\frac{1}{2}}\leq x^{\frac{1}{2}}\leq (\frac{36}{25})^{\frac{1}{2}}\\\\\frac{2}{3}\leq x^{\frac{1}{2}}\leq \frac{6}{5}\\\\\frac{2}{3}\leq x^{\frac{1}{2}}\leq 1\frac{1}{5}$