Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!! СРОЧНО!!!!!!!!!!!

Answer 1

$1)\; \; 9\sqrt2-\sqrt{98}=9\sqrt2-\sqrt{49\cdot 2}=9\sqrt2-7\sqrt2=2\sqrt2\approx 2,8\\\\2)\; \; \frac{\sqrt{15}+\sqrt5}{\sqrt5}=\frac{\sqrt{5\cdot 3}+\sqrt5}{\sqrt5} =\frac{\sqrt5\cdot (\sqrt3+1)}{\sqrt5}=\sqrt3+1\approx 2,7\\\\3)\; \; \frac{2\sqrt3-2}{\sqrt6-\sqrt2}=\frac{2\cdot (\sqrt3-1)}{\sqrt{2\cdot 3}-\sqrt2}=\frac{2\cdot (\sqrt3-1)}{\sqrt2\cdot (\sqrt3-1)}=\frac{2}{\sqrt2}=\sqrt2\approx 1,4$

$4)\; \; \frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}+\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}=\frac{(\sqrt5-\sqrt3)^2+(\sqrt5+\sqrt3)^2}{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)}=\frac{(8-2\sqrt{15})+(8+2\sqrt{15})}{(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)}=\\\\=\frac{8+8}{5-3}=\frac{16}{2}=8\\\\8>2\sqrt2>\sqrt3+1>\sqrt2$