Răspuns:
[tex]a=\sqrt{\frac{1}{\frac{1}{\sqrt{12}}-\frac{1}{\sqrt{27}}}*[(\frac{2}{\sqrt{3}+\sqrt{2}})^{-1} -\frac{1}{\sqrt{2}}}=\sqrt{\frac{1}{\frac{3)1}{2\sqrt{3}}-\frac{\2)1}{3\sqrt{3}}}*(\frac{\sqrt{3}+\sqrt{2}}{2} -\frac{\sqrt{2})1}{\sqrt{2}})}=\\=\sqrt{\frac{1}{\frac{1}{6\sqrt{3}}}*(\frac{\sqrt{3}+\sqrt{2}}{2} -\frac{\sqrt{2}}{2})}= \sqrt{6\sqrt{3}*\frac{\sqrt{3}+\sqrt{2}-\sqrt{2}}{2}}= \sqrt{6\sqrt{3}*\frac{\sqrt{3}}{2}}= \sqrt{3*3} =3[/tex]
Explicație pas cu pas:
