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ParthKohli

  • one year ago

If\[x = \dfrac{4ab}{a + b}\]Then prove that\[\dfrac{x + 2a}{x - 2a} + \dfrac{x + 2b}{x - 2b} = 2\]

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  1. ParthKohli
    • one year ago
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    @UnkleRhaukus

  2. ParthKohli
    • one year ago
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    I can't believe I'm finding 9th grade so hard. :'-|

  3. ParthKohli
    • one year ago
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    Ahahahahaha well! I got it.

  4. .Sam.
    • one year ago
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    lol

  5. UnkleRhaukus
    • one year ago
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    how?

  6. ParthKohli
    • one year ago
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    Do you want to know my proof?

  7. UnkleRhaukus
    • one year ago
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    is it elegant?

  8. ParthKohli
    • one year ago
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    \[x = \dfrac{4ab}{a+b} \qquad \Rightarrow \qquad \dfrac{x}{2a}=\dfrac{2ab}{a+b} ~~\text{and} ~~ \dfrac{x}{2b} = \dfrac{2a}{a+b}\]I don't know, it may be elegant.

  9. UnkleRhaukus
    • one year ago
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    i see what your doing there...

  10. ParthKohli
    • one year ago
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    OK, let me continue.\[\dfrac{x + 2a}{x - 2a} = \dfrac{2b + a + b}{2b - a - b}\]and\[\dfrac{x + 2b}{x-2b} = \dfrac{2a + a + b}{2a - a - b}\]by componendo and dividendo property

  11. ParthKohli
    • one year ago
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    Oh, and in my earlier post, I meant to write\[\dfrac{x}{2a} = \dfrac{2b}{a+b}\]

  12. ParthKohli
    • one year ago
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    Simplifying, we get\[\dfrac{x + 2a}{x - 2a} = \dfrac{3b+a}{b-a}\]and\[\dfrac{x + 2b}{x-2b} =\dfrac{3a+b}{a - b}\]

  13. ParthKohli
    • one year ago
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    So now we can add both equations

  14. ParthKohli
    • one year ago
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    \[\begin{aligned}\dfrac{x + 2a}{x - 2a} + \dfrac{x+2b}{x-2b} &=\dfrac{3b+a}{b-a} + \dfrac{3a+b}{a-b} \\ \\ \\ & = \dfrac{-3b-a+3a+b}{a-b} \\ \\ \\ &= \dfrac{2a-2b}{a-b} \\ \\ \\ & = 2\end{aligned} \]

  15. ParthKohli
    • one year ago
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    Phew. QED

  16. UnkleRhaukus
    • one year ago
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    this is 9th grade?

  17. ParthKohli
    • one year ago
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    Yeah.

  18. ParthKohli
    • one year ago
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    No wait, this is 8th grade. I was just reviewing stuff.

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