Media aritmetica a trei numere,y si z este egala cu 20 radicsl din 2.
a)Dacă media aritmetica a numerelor x si y este egala cu 3 radical din 2,aflati z.
b)Arătați ca radical din xplus y supra z € Q.

Răspuns :

[tex]\dfrac{x+y+z}{3} = 20\sqrt{2} \\ \\ \\ a)\,\,\,\, \dfrac{x+y}{2} = 3\sqrt{2} \Big|\cdot \dfrac{2}{3} \Rightarrow \dfrac{x+y}{3} = 2\sqrt{2}\Big|+\dfrac{z}{3} \Rightarrow[/tex]

[tex]\Rightarrow \dfrac{x+y+z}{3} = 2\sqrt{2}+\dfrac{z}{3} \Rightarrow 20\sqrt{2}=2\sqrt{2}+\dfrac{z}{3} \Big|\cdot 3\Rightarrow[/tex]

[tex]\Rightarrow 60\sqrt{2}=6\sqrt{2}+z \Rightarrow z = 60\sqrt{2}-6\sqrt{2}\Rightarrow[/tex]

[tex]\Rightarrow \boxed{z = 54\sqrt{2}}\\ \\ \\ b)\,\,\,\,\sqrt{\dfrac{x+y}{z}}= \sqrt{\dfrac{\dfrac{x+y}{2}}{\dfrac{z}{2}}} = \sqrt{\dfrac{3\sqrt{2}}{\dfrac{54\sqrt{2}}{2}}}= \sqrt{\dfrac{3\sqrt{2}}{27\sqrt{2}}} = \sqrt{\dfrac{3}{27}} = \\ = \sqrt{\dfrac{1}{9}} = \dfrac{1}{3} \in \mathbb{Q}[/tex]