Verified answer

[tex]\displaystyle\bf\\5.1\\\\\sqrt{x} =\frac{3}{7} \\\\\Big(\sqrt{x}\Big)^{2} =\Big(\frac{3}{7} \Big)\\\\x=\frac{9}{49} \\\\5.2\\\\\sqrt{x} =-4\\\\x\in\oslash[/tex]

корней нет

[tex]\displaystyle\bf\\5.3\\\\x^{2} =25\\\\x_{1,2} =\pm \ \sqrt{25} =\pm \ 5\\\\x_{1} =-5 \ \ ; \ \ x_{2} =5\\\\5.4\\\\x^{2} =-9\\\\x\in\oslash[/tex]

корней нет

[tex]\displaystyle\bf\\6.1\\\\\frac{x^{2} -11}{x-\sqrt{11} } =\frac{x^{2}-(\sqrt{11} )^{2} }{x-\sqrt{11} } =\frac{(x-\sqrt{11} )\cdot(x+\sqrt{11} )}{x-\sqrt{11} } =x+\sqrt{11} \\\\6.2\\\\\frac{7\sqrt{2} -2}{5\sqrt{2} } =\frac{7\sqrt{2} -(\sqrt{2} )^{2} }{5\sqrt{2} } =\frac{\sqrt{2}\cdot(7-\sqrt{2} ) }{5\sqrt{2} } =\frac{7-\sqrt{2} }{5}[/tex]