Question

3 Calculaţi:

cu rezolvare va rog dacă se poate​

Answer 1

$\displaystyle a)~\frac{1}{\sqrt{2} } +\frac{^{\sqrt{2}+1) }1}{\sqrt{2} -1} =\frac{1}{\sqrt{2} } +\frac{\sqrt{2} +1}{\Big(\sqrt{2} -1\Big)\Big(\sqrt{2} +1\Big)} =\frac{1}{\sqrt{2} } +\frac{\sqrt{2} +1}{2-1} =\\\\=\frac{1}{\sqrt{2} }+^{\sqrt{2}) }\sqrt{2} +^{\sqrt{2} )}1= \frac{1+2+\sqrt{2} }{\sqrt{2} } =\frac{3+\sqrt{2} }{\sqrt{2} }$

$\displaystyle b)~\frac{^{\sqrt{3}-1) }1}{\sqrt{3}+1 } +\frac{^{3+\sqrt{3}) }1}{3-\sqrt{3} }-\frac{1}{\sqrt{3} }=\\\\= \frac{\sqrt{3}-1 }{\Big(\sqrt{3}+1\Big)\Big(\sqrt{3} -1\Big) }+\frac{3+\sqrt{3} }{\Big(3-\sqrt{3}\Big)\Big(3+\sqrt{3} \Big) } -\frac{1}{\sqrt{3} }=\\\\=\frac{\sqrt{3}-1 }{3-1} +\frac{3+\sqrt{3} }{9-3} -\frac{1}{\sqrt{3} }=\frac{^{3\sqrt{3}) }\sqrt{3} -1}{2} +\frac{^{\sqrt{3}) }3+\sqrt{3} }{6} -\frac{^{6) }1}{\sqrt{3} } =$

$\displaystyle =\frac{3\sqrt{3}\Big( \sqrt{3} -1\Big)+\sqrt{3}\Big(3+\sqrt{3}\Big)-6 }{6\sqrt{3} } =\frac{9-3\sqrt{3}+3\sqrt{3} +3-6 }{6\sqrt{3} } =\frac{6}{6\sqrt{3} } =\frac{1}{\sqrt{3} }$

$\displaystyle c)~\frac{18}{\sqrt{6} }-\frac{^{\sqrt{6}+1) }5}{\sqrt{6}-1 } -\sqrt{24}=\frac{18}{\sqrt{6} } - \frac{5\Big(\sqrt{6} +1\Big)}{\Big(\sqrt{6} -1\Big)\Big(\sqrt{6}+1\Big) } -2\sqrt{6} =\\\\=\frac{18}{\sqrt{6} } -\frac{5\sqrt{6} +5}{6-1} -2\sqrt{6} =\frac{^{5)}18}{\sqrt{6} }-\frac{^{\sqrt{6}) }5\sqrt{6}+5 }{5} -^{5\sqrt{6}) }2\sqrt{6}=\\\\=\frac{5 \cdot 18-\sqrt{6}\Big(5\sqrt{6}+5\Big)-5\sqrt{6}\cdot 2\sqrt{6}}{5\sqrt{6}}=\frac{90-30-5\sqrt{6} -60}{5\sqrt{6} } =-\frac{5\sqrt{6} }{5\sqrt{6} }=-1$