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E(x)=[(2x+1)/(3x-2)]/[1-(5x^2-3)/(9x^2-4)].Is it E(x)=(3x+2)/(2x-1)? Find x integer so...
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First of all, when a fraction is divided by the other, in fact we'll multiply one of them with the inverse of the other one. For example:
In our expression E(x), before applying the rule above, we have to have the same denominator in the expression [1-(5x^2-3)/(9x^2-4)].
We've noticed that at the numerator and at the denominator, we have difference of squares, like:
Now we can apply the rule:
We can reduce the similar terms (2x+1) and (3x-2).
For finding x integer, in order to obtain E(x) natural, (3x+2) has to be divided by (2x-1).
We'll write the ule of division with reminder:
If we'd have
E(x)= (3x+2)/(2x-1)=3/2+7/2=10/2=5, which is a natural number
For this, 2x-1=1, 2x=2, x=1, which is integer
Posted by giorgiana1976 on June 2, 2009 at 3:26 AM (Answer #1)
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