Răspuns :

100000000000000000id

Răspuns:

[tex] \frac{1}{2} = 1 - \frac{1}{2} [/tex]

[tex] \frac{1}{2 \times 3} = \frac{1}{2} - \frac{1}{3} [/tex]

[tex] \frac{1}{3 \times 4} = \frac{1}{3} - \frac{1}{4} [/tex]

[tex]...[/tex]

[tex] \frac{1}{x {}^{2 } + x } = \frac{1}{x(x + 1)} = \frac{1}{x} + \frac{1}{x + 1} = > 1 - \frac{1}{x+ 1} = \frac{2018}{2019} = > \frac{x}{x + 1} = \frac{2018}{2019} = > 2018x + 2018 = 2019x = > x = 2018[/tex]

Mult succes în continuare!

Targoviste44id

[tex]\it \dfrac{1}{x^2+x} =\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}[/tex]

Suma se poate scrie:

[tex]\it \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ ...\ +\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2018}{2019}[/tex]

După multele reduceri de termeni opuși, se obține:

[tex]\it 1-\dfrac{1}{x+1}=\dfrac{2018}{2019} \Rightarrow \dfrac{x+1-1}{x+1}=\dfrac{2018}{2019} \Rightarrow \dfrac{x}{x+1}=\dfrac{2018}{2019} \Rightarrow x=2018[/tex]

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