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Va rog ajutorr
Se considera expresia E(x)= x supra (x^2 + x)-( x supra x-1 - x supra x+1): 2x supra x-1 . Aratati ca E(x)=0


Răspuns :

Răspuns:

[tex] \frac{x}{ {x}^{2} + x } - (\frac{x}{x - 1} - \frac{x}{x + 1} ) \div \frac{2x}{x - 1} [/tex]

[tex] \frac{ x}{x(x + 1)} - \frac{x(x + 1) - x(x - 1)}{(x + 1)( x - 1)} \times \frac{x - 1}{2x} [/tex]

[tex] \frac{1}{x + 1} - \frac{x(x + 1 - x + 1)}{x + 1} \times \frac{1}{2x} [/tex]

[tex] \frac{1}{x + 1} - \frac{2x}{x + 1} \times \frac{1}{2x} [/tex]

[tex] \frac{1}{x + 1} - \frac{1}{x + 1} [/tex]

[tex] \frac{1 - 1}{x + 1} [/tex]

[tex] \frac{0}{x + 1} [/tex]

[tex]0[/tex]