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Simplify the following: a) (3u+2v+3wu+2wv) / (3wv+3u+2v)

Mathematics

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Answer

(3u+2v)(1+w)3u(v+1)+2v(w+1)\frac{(3u + 2v)(1 + w)}{3u(v + 1) + 2v(w + 1)}

Step-by-step solution

3 steps

Huu hapa ni urahisishaji wa usemi uliotolewa:

  1. Tafuta vigawo vya nambari.

    Nambari ni 3u+2v+3wu+2wv3u + 2v + 3wu + 2wv.
    Tupange upya maneno na kutoa vigawo vya kawaida:

    (3u+3wu)+(2v+2wv)(3u + 3wu) + (2v + 2wv)

    Toa 3u3u kutoka kundi la kwanza na 2v2v kutoka kundi la pili:

    3u(1+w)+2v(1+w)3u(1 + w) + 2v(1 + w)

    Toa kigawo cha kawaida (1+w)(1 + w):

    (3u+2v)(1+w)(3u + 2v)(1 + w)
  2. Tafuta vigawo vya denomineta.

    Denomineta ni 3uv+3u+2vw+2v3uv + 3u + 2vw + 2v.
    Tupange upya maneno na kutoa vigawo vya kawaida:

    (3uv+3u)+(2vw+2v)(3uv + 3u) + (2vw + 2v)

    Toa 3u3u kutoka kundi la kwanza na 2v2v kutoka kundi la pili:

    3u(v+1)+2v(w+1)3u(v + 1) + 2v(w + 1)
  3. Andika usemi uliorahisishwa.

    Sasa tunaweza kuandika upya sehemu kwa kutumia vigawo tulivyopata:

    (3u+2v)(1+w)3u(v+1)+2v(w+1)\frac{(3u + 2v)(1 + w)}{3u(v + 1) + 2v(w + 1)}

    Hakuna vigawo vya kawaida kati ya nambari na denomineta ambavyo vinaweza kufutwa. Kwa hivyo, huu ndio mfumo uliorahisishwa zaidi.

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