Question

patru colegi, Maria,Monica,Ionel,Antim au calculat media aritmetica a numerelelor a=rad(4rad3-8)² si b=|4V3+8|
Va rog ajutatima​

Answer 1

$\it a=\sqrt{(4\sqrt3-8)^2}=|4\sqrt3-8|\\ \\ 4\sqrt3=\sqrt{4^2\cdot3}=\sqrt{16\cdot3}=\sqrt{48} < \sqrt{64}=8 \Rightarrow4\sqrt3 < 8 \Rightarrow\\ \\ \Rightarrow 4\sqrt3-8 < 0 \Rightarrow |4\sqrt3-8|=-(4\sqrt3-8)=-4\sqrt3+8\\ \\ a=-4\sqrt3+8\\ \\ b=|4\sqrt3+8|=4\sqrt3+8,\ deoarece\ 4\sqrt3+8 > 0\\ \\ \\ m_a=\dfrac{a+b}{2}=\dfrac{-4\sqrt3+8+4\sqrt3+8}{2}=\dfrac{16}{2}=8$