Răspuns :
[tex]\sqrt{(3-\sqrt{10})^2}+\sqrt2(2\sqrt2-\sqrt5)=\\=\sqrt{10}-3+\sqrt2 \cdot \sqrt8-\sqrt2 \cdot \sqrt5=\\=\sqrt{10}-3+\sqrt{16}-\sqrt{10}= \\= -3+4 =\boxed{1} \to \bold{num\u{a}r \ intreg}[/tex]
[tex]\sqrt{(3-\sqrt{10})^2}+\sqrt2(2\sqrt2-\sqrt5)=\\=\sqrt{10}-3+\sqrt2 \cdot \sqrt8-\sqrt2 \cdot \sqrt5=\\=\sqrt{10}-3+\sqrt{16}-\sqrt{10}= \\= -3+4 =\boxed{1} \to \bold{num\u{a}r \ intreg}[/tex]