Resposta:
[tex]x^2-\frac{1}{3}=\frac{1}{6}x^2\\\\x^2-\frac{1}{3}+\frac{1}{3}=\frac{1}{6}x^2+\frac{1}{3}\\\\x^2=\frac{1}{6}x^2+\frac{1}{3}\\\\x^2-\frac{1}{6}x^2=\frac{1}{6}x^2+\frac{1}{3}-\frac{1}{6}x^2\\\\\frac{5}{6}x^2=\frac{1}{3}\\\\6\cdot \frac{5}{6}x^2=\frac{1\cdot \:6}{3}\\\\5x^2=2\\\\\frac{5x^2}{5}=\frac{2}{5}\\\\x^2=\frac{2}{5}\\\\x=\sqrt{\frac{2}{5}},\:x=-\sqrt{\frac{2}{5}}[/tex]
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[tex]\frac{x^2}{4}+\frac{1}{10}=\frac{x^2}{5}+\frac{x}{2}\\\\\frac{x^2}{4}\cdot \:20+\frac{1}{10}\cdot \:20=\frac{x^2}{5}\cdot \:20+\frac{x}{2}\cdot \:20\\\\5x^2+2=4x^2+10x\\\\5x^2+2-10x=4x^2+10x-10x\\\\5x^2+2-10x=4x^2\\\\5x^2+2-10x-4x^2=4x^2-4x^2\\\\x^2-10x+2=0\\\\x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}\\\\x_{1,\:2}=\frac{-\left(-10\right)\pm \:2\sqrt{23}}{2\cdot \:1}\\\\[/tex]
[tex]x_1=\frac{-\left(-10\right)+2\sqrt{23}}{2\cdot \:1},\:x_2=\frac{-\left(-10\right)-2\sqrt{23}}{2\cdot \:1}\\\\x=5+\sqrt{23}\\\\\:x=5-\sqrt{23}[/tex]
