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Kris5

  • 3 years ago

1/x^2+4x+3+ 1/ x^2-1 ...please help me solve this

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  1. nphuongsun93
    • 3 years ago
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    is it this\[\huge \frac{1}{x^2} +4x + 3 + \frac{1}{x^2-1}\]?

  2. Kris5
    • 3 years ago
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    no the 4x+3 goes under 1 in the first fraction

  3. nphuongsun93
    • 3 years ago
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    \[\frac{1}{x^2+4x+3}+ \frac{1}{x^2-1}\]right?

  4. Kris5
    • 3 years ago
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    yes :)

  5. nphuongsun93
    • 3 years ago
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    let's see what i can do with this, just simplify i guess first factor \[x^2 + 4x + 3 \space and \space x^2 - 1\]

  6. Kris5
    • 3 years ago
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    (x+3) (x+1) and (x+1) (x-1)

  7. nphuongsun93
    • 3 years ago
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    \[\frac{1}{(x+3)(x+1)} + \frac{1}{(x+1)(x-1)}\]make the denominators common~

  8. Kris5
    • 3 years ago
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    that's what I don't get..I'm stuck there

  9. nphuongsun93
    • 3 years ago
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    \[\frac{(x-1)+(x+3)}{(x+3)(x+1)(x-1)}\]rings any bells? :s

  10. Kris5
    • 3 years ago
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    sort of

  11. nphuongsun93
    • 3 years ago
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    |dw:1348609884918:dw|

  12. Kris5
    • 3 years ago
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    okay

  13. nphuongsun93
    • 3 years ago
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    \[\frac{2x+2}{(x+3)(x+1)(x-1)}=\frac{2(x+1)}{(x+3)(x+1)(x-1)}=\frac{2}{(x+3)(x-1)}\]

  14. Kris5
    • 3 years ago
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    thanks :)

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