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Sebastiann
Pe: 2013-02-03T11:06:32+02:00 In: In:

Expresii

Am de facut o expresie si nu ii pot da de cap.. Ma poate ajuta cineva?Va rog 🙂 🙂
E(x) = ( 2/x-3 -1/x+1) :9xpatrat -1/(x-3)(x+1) + (x+2)(x+1)-xpatrat +5/3x+1

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Similare

5 raspunsuri

  1. mihai2000

    \frac{\frac{2}{x-3}+\frac{1}{x+1}}{\frac{9{x}^{2}-1}{(x-3)(x+1)}}+\frac{(x+2)(x+1)-{x}^{2}+5}{3x+1}

    Aduci la acelasi numitor numaratorul primei fractii
    Simplifici cu (x-3)(x+1), tii cont ca 9x^2-1 = (3x-1)(3x+1), din (a^2-b*2) =(a-b)(a+b)

    \frac{2x+2+x-3}{(3x-1)(3x+1)}+\frac{{x}^{2}+3x+2-{x}^{2}+5}{3x+1}
    Mai faci niste calcule …

  2. Sebastiann

    II bine?:)

    Attached files

  4. mihai2000

    Prima fractie (la care am ajuns, vezi mai sus) se simplifica cu 3x+1 (daca se fac calcule la numarator), din a doua dispare x^2 …

  5. dennis9091
    guru (IV)

    Sebastiann descompune numaratorul cu ecuatia de gradul 2 sau cu schema horner.. cu ti’e mai usor.

  6. edy8
    maestru (V)

    \it{\bl \Large I) \frac{2}{x-3}+\frac{1}{x+1} = \frac{2x+2+x-3}{(x-3)(x+1)} = \frac{3x-1}{(x-3)(x+1)}\\\;\\II) \frac{9x^2-1}{(x-3)(x+1)} = \frac{(3x)^2-1}{(x-3)(x+1)} = \frac{(3x-1)(3x+1)}{(x-3)(x+1)}\\\;\\III) (\frac{2}{x-3}+\frac{1}{x+1}) : \frac{9x^2-1}{(x-3)(x+1)} = \frac{3x-1}{(x-3)(x+1)} : \frac{(3x-1)(3x+1)}{(x-3)(x+1)} = \frac{3x-1}{(x-3)(x+1)} \cdot \frac{(x-3)(x+1)}{(3x-1)(3x+1)} = \frac{1}{3x+1}.\\\;\\IV) \frac{(x+2)(x+1)-x^2+5}{3x+1} = \frac{x^2+3x+2-x^2+5}{3x+1} = \frac{3x+7}{3x+1}\\\;\\E(x) = \frac{1}{3x+1} + \frac{3x+7}{3x+1} =\frac{3x+8}{3x+1} = 1+ \frac{7}{3x+1}.}

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