Question

m Aritmetic, media geometrica, madia inaltimi x=radical din 3+1,y=radical din 3-1​

Answer 1

$x=\sqrt3+1; \\ y=\sqrt{3}-1; \\ M_A=\frac{x+y}{2}=\frac{\sqrt3+1+\sqrt3-1}{2}=\frac{2\sqrt3}{2}=\boxed{\sqrt{3}}; \\ M_G=\sqrt{x\cdot y}=\sqrt{(\sqrt3+1)(\sqrt3-1)}=\sqrt{(\sqrt3)^2-1^2}=\sqrt{3-1}=\boxed{\sqrt{2}}.$