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

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ParthKohliBest ResponseYou've already chosen the best response.1
I can't believe I'm finding 9th grade so hard. :'
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Ahahahahaha well! I got it.
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Do you want to know my proof?
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[x = \dfrac{4ab}{a+b} \qquad \Rightarrow \qquad \dfrac{x}{2a}=\dfrac{2ab}{a+b} ~~\text{and} ~~ \dfrac{x}{2b} = \dfrac{2a}{a+b}\]I don't know, it may be elegant.
 11 months ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i see what your doing there...
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
OK, let me continue.\[\dfrac{x + 2a}{x  2a} = \dfrac{2b + a + b}{2b  a  b}\]and\[\dfrac{x + 2b}{x2b} = \dfrac{2a + a + b}{2a  a  b}\]by componendo and dividendo property
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Oh, and in my earlier post, I meant to write\[\dfrac{x}{2a} = \dfrac{2b}{a+b}\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
Simplifying, we get\[\dfrac{x + 2a}{x  2a} = \dfrac{3b+a}{ba}\]and\[\dfrac{x + 2b}{x2b} =\dfrac{3a+b}{a  b}\]
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
So now we can add both equations
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[\begin{aligned}\dfrac{x + 2a}{x  2a} + \dfrac{x+2b}{x2b} &=\dfrac{3b+a}{ba} + \dfrac{3a+b}{ab} \\ \\ \\ & = \dfrac{3ba+3a+b}{ab} \\ \\ \\ &= \dfrac{2a2b}{ab} \\ \\ \\ & = 2\end{aligned} \]
 11 months ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
this is 9th grade?
 11 months ago

ParthKohliBest ResponseYou've already chosen the best response.1
No wait, this is 8th grade. I was just reviewing stuff.
 11 months ago
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