Question

\frac{1}{3\sqrt{3}+4}-(\frac{1}{\sqrt{2}} -\frac{1}{2} ):\frac{/1-\sqrt{2}/}{\sqrt{18}}+\sqrt{9}
calculati

Answer 1

   
$\displaystyle\\ \frac{1}{3\sqrt{3}+4}-\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right):\frac{|1-\sqrt{2}|}{\sqrt{18}}+\sqrt{9}=\\\\\\ =\frac{1}{3\sqrt{3}+4}-\left(\frac{1}{\sqrt{2}}-\frac{1}{2}\right):\frac{\sqrt{2}-1}{3\sqrt{2}}+\sqrt{9}=~\text{(Rationalizam numitorii.)}\\\\\\ =\frac{3\sqrt{3}-4}{27-16}-\left(\frac{\sqrt{2}}{2}-\frac{1}{2}\right):\frac{\sqrt{2}(\sqrt{2}-1)}{6}+\sqrt{9}=\\\\\\=\frac{3\sqrt{3}-4}{11}-\frac{\sqrt{2}-1}{2}:\frac{\sqrt{2}(\sqrt{2}-1)}{6}+\sqrt{9}=$


[tex]\displaystyle\\ =\frac{3\sqrt{3}-4}{11}-\frac{\sqrt{2}-1}{2}\times\frac{6}{\sqrt{2}(\sqrt{2}-1)}+\sqrt{9}= \\\\\\ =\frac{3\sqrt{3}-4}{11}-\frac{6(\sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}+\sqrt{9}=\\\\\\ =\frac{3\sqrt{3}-4}{11}-\frac{3}{\sqrt{2}}+\sqrt{9}=\\\\\\ =\frac{3\sqrt{3}-4}{11}-\frac{3\sqrt{2}}{2}+3=~\text{(Aducem la acelasi numitor.)}\\\\\\ =\frac{2(3\sqrt{3}-4)}{2\times11}-\frac{11\times3\sqrt{2}}{11\times2}+\frac{22\times3}{22}=\\\\\\ [/tex]


[tex]\displaystyle\\ =\frac{6\sqrt{3}-8}{22}-\frac{33\sqrt{2}}{22}+\frac{66}{22}=\\\\\\ =\frac{6\sqrt{3}-8 -33\sqrt{2}+66}{22}=\boxed{\frac{58+6\sqrt{3}-33\sqrt{2}}{22}}[/tex]