Question

Determinati valoarea de adevar a propozitiilor:

a) x ∈ R
b) x ∈ Q
c) x ∈ R \ Q
d) x ∈ Z

unde x= $\sqrt{2010+2 ori (1+2+3+...+2009)}$ .

Answer 1

Răspuns:

Explicație pas cu pas:

1+2+3+....+n=n(n+1)/2

1+2+...+2009=2009(2009+1)/2=2009x2010

x=$\sqrt{2010+2*2009*2010/2} =\sqrt{2010+2009*2010} =\sqrt{2010( 1+2009)=\sqrt{2010*2010} =2010$

a. A, b. A, c. F, d A