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Justmeale2345
a fost răspuns

se se efectueze:
[tex](2013 \times ( \frac{1}{2012} \times (\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \frac{1}{4 \times 5} + ... + \frac{1}{2012 \times 2013} ))) - 1 = [/tex]
Va rog sa imi si explicati.....exerciutul l-am facut in clasa dar nu stiu daca am copiat chiar bine rezolvarea si nici nu am inteles cum se face

Răspuns :

Targoviste44

[tex]\it \dfrac{1}{1\cdot2} +\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\ ...\ +\dfrac{1}{2012\cdot2013} =\\ \\ \\=\dfrac{1}{1}-\dfrac{1}{2} +\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ ...\ +\dfrac{1}{2012}-\dfrac{1}{2013}=1-\dfrac{1}{2013}=\dfrac{2012}{2013}[/tex]

Expresia devine:

[tex]\it 2013\cdot\dfrac{1}{2012}\cdot\dfrac{2012}{2013}-1=1-1=0[/tex]


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