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$\begingroup$

$$\frac{h}{2xh+h^2} = \frac{h}{h(2x+h)} = \frac{1}{2x+h}$$

I do not understand who the 1 comes to place can anyone give me some clarity ill jsut get lost

Same at this one $$\frac{3^3*3^{2x}+3^{2x}}{3^2*3^x-3^x} = \frac{3^{2x}(3^{3}+3^{2x}}{3^x(3^2-3^x)} = \frac{3x*7}{2}$$ but i still dont understand how it is 27+1/9-1

I understand h/h=1 and so on but x/xX+xX is 1/2x i dont understand this really... Thanks in forehand

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    Hint: $\;\cfrac{a}{b} = a \cdot \cfrac{1}{b}\;$ and $\;\cfrac{a}{b\,c} = \cfrac{a}{b}\cdot \cfrac{1}{c}\;$. – 2017-02-05
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    this does not tell me alot, in second expression how does 3^2x/3^x become a 1? – 2017-02-05
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    Take $\;a=h, b=h, c=2x+h\;$ in the $2^{nd}$ identity and see what you get. P.S. I edited your question and formatted the first equalities using [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Please format the rest of your post using similar syntax, since it's hard to read as it is now. – 2017-02-05
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    yes i will, ty didnt know how i did – 2017-02-05
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    $\frac {3^{2x}}{3^x}$ doe not become a 1. It becomes a $3^x$. Because $\frac{3^{2x}}{3^{x}} = \frac{(3^x)^2}{3^x} = 3^x$. – 2017-02-05
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    You are simply canceling out common factors. – 2017-02-05
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    I don't like to edit your post because I don't want to second guess your meaning but do you mean: $\frac{3^3*3^{2x}+3^{2x}}{3^2*3^x-3^x} = \frac{3^{2x}(3^{3}+1)}{3^x(3^2-1)} = \frac{3x*7}{2}$? Because what you have right now is simply wrong. – 2017-02-05
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    yes i knew something was wrong, my intention was to show were my brain did not understand. but i still quite dont understand why do u get 1 on both sides? of the fraction. is it simply becuase 3/3=1/1? – 2017-02-07

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$\frac{h}{2xh+h^2} = \frac{h}{h(2x+h)} =\frac{\not h}{\not h(2x+h)}=\frac{1}{2x+h}$

and

$\frac{3^3*3^{2x}+3^{2x}}{3^2*3^x-3^x} = \frac{3^{2x}(3^{3}+1)}{3^x(3^2-1)} = \frac{(3^x)^2(28)}{3^x(8)}=\frac{\not 3^x3^x*\not 4*7}{\not 3^x*\not 4*2}=\frac{3^x*7}{2}$