Question

Dau 80 pentru evaluare

Answer 1

$\it 2)\ 2\sqrt7=\sqrt{2^2\cdot7}=\sqrt{28}\\ \\ \sqrt3<\sqrt{28} \Rightarrow \sqrt3<2\sqrt7 \Rightarrow \sqrt3-2\sqrt7<0 \Rightarrow |3-2\sqrt7|=-3+2\sqrt7\\ \\ \\ 3)\ =32\cdot(-82,65)+4\cdot2,2 =-2644,8+8,8=-2636\\ \\ \\ 4)\ (-\infty,\ -2\sqrt3)\ \cup\ [-\sqrt7,\ 50] = (-\infty,\ 50]\backslash [-2\sqrt3,\ -\sqrt7)$

$\it 5)\ 3(x+7) =21|_{:3} \Rightarrow x+7=7 \Rightarrow x=0\in\mathbb{Z}$