Question

Va rog frumos ......​

Answer 1

$\displaystyle a=\Bigg[\Bigg(\sqrt{8} +\frac{\sqrt{2} }{5} -\frac{\sqrt{18} }{2} \Bigg)-\Bigg(\frac{3\sqrt{8} }{\sqrt{16} } +\frac{2\sqrt{18} }{3} \Bigg)\Bigg]:\sqrt{\frac{2}{25} } =\\\\=\Bigg[\Bigg(~^{10)}2\sqrt{2} +\frac{^{2)}\sqrt{2} }{5} -\frac{^{5)}3\sqrt{2} }{2} \Bigg)-\Bigg(\frac{3 \cdot 2\sqrt{2} }{4} +\frac{2 \cdot 3\sqrt{2} }{3} \Bigg)\Bigg] :\frac{\sqrt{2} }{\sqrt{25} } =$

$\displaystyle =\Bigg[\frac{10 \cdot 2\sqrt{2}+2\sqrt{2} -5 \cdot3\sqrt{2} }{10} -\Bigg(\frac{^{3)}6\sqrt{2} }{4} +\frac{^{4)}6\sqrt{2} }{3} \Bigg)\Bigg]:\frac{\sqrt{2} }{5} =\\\\=\Bigg(\frac{20\sqrt{2}+2\sqrt{2}-15\sqrt{2}}{10} -\frac{3\cdot 6\sqrt{2}+4 \cdot 6\sqrt{2} }{12} \Bigg) \cdot \frac{5}{\sqrt{2} } =\\\\ =\Bigg(\frac{7\sqrt{2} }{10} -\frac{18\sqrt{2}+24\sqrt{2}}{12}\Bigg) \cdot \frac{5} {\sqrt{2} } =\Bigg(\frac{^{12)}7\sqrt{2} }{10} -\frac{^{10)}42\sqrt{2} }{12} \Bigg)\cdot \frac{5}{\sqrt{2} }=$

$\displaystyle =\frac{12 \cdot 7\sqrt{2} -10 \cdot 42\sqrt{2} }{120} \cdot \frac{5}{\sqrt{2} } =\frac{84\sqrt{2} -420\sqrt{2} }{120} \cdot \frac{5}{\sqrt{2} }= -\frac{336\sqrt{2} }{120} \cdot \frac{5}{\sqrt{2} } =\\\\=-\frac{1680\sqrt{2} }{120\sqrt{2} } =-\frac{1680}{120}= -14 \in\mathbb{Z}$