Odpowiedź:
[tex]\frac{x+\sqrt{2}}{\sqrt{3}+1}<\frac{x+\sqrt{3}}{\sqrt{2}+1}\ \ /\cdot(\sqrt{3}+1)(\sqrt{2}+1)\\\\(\sqrt{2}+1)(x+\sqrt{2})<(\sqrt{3}+1)(x+\sqrt{3})\\\\\sqrt{2}x+2+x+\sqrt{2}<\sqrt{3}x+3+x+\sqrt{3}\\\\\sqrt{2}x+x-\sqrt{3}x-x<3+\sqrt{3}-2-\sqrt{2}\\\\\sqrt{2}x-\sqrt{3}x<1+\sqrt{3}-\sqrt{2}\\\\(\sqrt{2}-\sqrt{3})x<1+\sqrt{3}-\sqrt{2}\ \ /:(\sqrt{2}-\sqrt{3})\\\\x>\frac{1+\sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}[/tex]
[tex]x>\frac{1+\sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}\cdot\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}\\\\x>\frac{(1+\sqrt{3}-\sqrt{2})(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}\\\\x>\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+3-2-\sqrt{6}}{(\sqrt{2})^2-(\sqrt{3})^2}\\\\x>\frac{\sqrt{2}+\sqrt{3}+1}{2-3}\\\\x>\frac{\sqrt{2}+\sqrt{3}+1}{-1}\\\\x>-(\sqrt{2}+\sqrt{3}+1)\\\\x>-\sqrt{2}-\sqrt{3}-1[/tex]
