Dla ułatwienia zadania najpierw wyliczymy wartość x podniesioną do kwadratu.
a)
[tex]x = \sqrt{2}-1\\\\(a-b)^2=a^2-2ab+b^2\\ x^2 = (\sqrt{2}-1)^2=2-2\sqrt{2}+1 \\x^2=3-2\sqrt{2} \\\\\frac{x^2}{x^2-3} =\frac{3-2\sqrt{2}}{3-2\sqrt{2}-3} =\frac{3-2\sqrt{2}}{-2\sqrt{2} } *\frac{\sqrt{2} }{\sqrt{2} }=\frac{3\sqrt{2} -2\sqrt{2}*\sqrt{2} }{-2\sqrt{2}*\sqrt{2} }= \\\\=\frac{3\sqrt{2} -4}{-4} =1-\frac{3}{4} \sqrt{2}[/tex]
b)
[tex]x=\sqrt{3}- \sqrt{2} \\x^2=3-2*\sqrt{3} *\sqrt{2} +2=5-2\sqrt{6} \\\frac{x^2}{x^2-3} =\frac{5-2\sqrt{6}}{5-2\sqrt{6}-3} = \frac{5-2\sqrt{6}}{2-2\sqrt{6}} *\frac{2+2\sqrt{6}}{2+2\sqrt{6}} =\\\\=\frac{(5-2\sqrt{6} )(2+2\sqrt{6})}{(2-2\sqrt{6})(2+2\sqrt{6})} \\\\(a+b)(a-b)=a^2-b^2\\\\\frac{(5-2\sqrt{6} )(2+2\sqrt{6})}{(2-2\sqrt{6})(2+2\sqrt{6})} = \frac{10+10\sqrt{6} -4\sqrt{6} -24}{4-24} =\frac{6\sqrt{6} -14}{-20} =\frac{7-3\sqrt{6} }{10}[/tex]
