Wyznacz liczbę odwrotną do liczby a:
1) [tex]a= \frac{3}{\sqrt{2}+1 } -\frac{9\sqrt{2}+4 }{3}
2) a = \frac{\sqrt{3} }{\sqrt{2}-\sqrt{3} } - \frac{1-5\sqrt{6} }{5}[/tex]
1) [tex]a= \frac{3}{\sqrt{2}+1 } -\frac{9\sqrt{2}+4 }{3}
2) a = \frac{\sqrt{3} }{\sqrt{2}-\sqrt{3} } - \frac{1-5\sqrt{6} }{5}[/tex]
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1)
[tex]a=\frac{3}{\sqrt2+1}-\frac{9\sqrt2+4}{3}=\frac{3}{\sqrt2+1}*\frac{\sqrt2-1}{\sqrt2-1}-\frac{9\sqrt2+4}{3}=\frac{3\sqrt2-3}{2-1}-\frac{9\sqrt2+4}{3}=\frac{3\sqrt2-3}{1}-\frac{9\sqrt2+4}{3}=\\\\=\frac{9\sqrt2-9}{3}-\frac{9\sqrt2+4}{3}=\frac{9\sqrt2-9-9\sqrt2-4}{3}=\frac{-13}{3}=-\frac{13}{3}\\\\\frac{1}{a}=\frac{1}{-\frac{13}{3}}=-\frac{3}{13}[/tex]
2)
[tex]a=\frac{\sqrt3}{\sqrt2-\sqrt3}-\frac{1-5\sqrt6}{5}=\frac{\sqrt3}{\sqrt2-\sqrt3}*\frac{\sqrt2+\sqrt3}{\sqrt2+\sqrt3}-\frac{1-5\sqrt6}{5}=\frac{\sqrt6+3}{2-3}-\frac{1-5\sqrt6}{5}=\\\\=\frac{\sqrt6+3}{-1}-\frac{1-5\sqrt6}{5}=\frac{-5\sqrt6-15}{5}-\frac{1-5\sqrt6}{5}=\frac{-5\sqrt6-15-1+5\sqrt6}{5}=\frac{-16}{5}=-\frac{16}{5}\\\\\frac{1}{a}=\frac{1}{-\frac{16}{5}}=-\frac{5}{16}[/tex]