Question

Mersiiiiiii multttt de toooot !
1 si 2 va rooog

Answer 1

Sper ca este bine....

Answer 2

    
[tex]1) \\ \text{Consider ca primul termen este: } \sqrt{( \sqrt{3}-1)^2} ~in ~loc~ de~ \sqrt{( \sqrt{3-1})^2}. \\ \\ \sqrt{( \sqrt{3}-1)^2} +\sqrt{( \sqrt{3}-3)^2}= \\ =( \sqrt{3}-1)+( \sqrt{3}-3)=\sqrt{3}-1+ \sqrt{3}-3 = \boxed{2\sqrt{3}-4}\Rightarrow R: \boxed{B} [/tex]


[tex]2) \\ \displaystyle \frac{2013}{ \frac{1}{1 \times 2} + \frac{1}{2 \times 3}+\hdots + \frac{1}{2013 \times 2014}}= \\ \\ =\frac{2013}{ \frac{1}{1}- \frac{1}{2} +\frac{1}{2}- \frac{1}{3}+\hdots + \frac{1}{2013}-\frac{1}{ 2014}}= \\ \\ \\ =\frac{2013}{ \frac{1}{1}-\frac{1}{ 2014}}=\frac{2013}{ \frac{2014}{2014}-\frac{1}{ 2014}}=\frac{2013}{ \frac{2013}{2014}}=2013 \times \frac{2014}{2013}= \boxed{2014}\Rightarrow R: \boxed{B}[/tex]