0;x>0\\\a+\frac{1}{a}=2.5\\2a^2-5a+2=0\\a_{1,2}=\frac{5^+_-3}{4}\\a_1=2;a_2=\frac{1}{2}\\x_1=4;x_2=\frac{1}{4}" alt="\sqrt{x}+\frac{1}{\sqrt{x}}=2,5\\\sqrt{x}=a;a>0;x>0\\\a+\frac{1}{a}=2.5\\2a^2-5a+2=0\\a_{1,2}=\frac{5^+_-3}{4}\\a_1=2;a_2=\frac{1}{2}\\x_1=4;x_2=\frac{1}{4}" align="absmiddle" class="latex-formula">