MHB Simplifying compound fractions

AI Thread Summary
To simplify the compound fraction [1/(1+x+h) - 1/(1+x)] / h, finding a common denominator for the top two fractions is essential. The common denominator is (1+x+h)(1+x), allowing for the expression to be rewritten as [(1+x) - (1+x+h)] / [(1+x+h)(1+x)]. This results in a simplified numerator of -h, leading to the overall simplification of the compound fraction. The final expression can be further reduced by dividing by h, yielding -1/[(1+x+h)(1+x)]. Understanding how to manipulate common denominators is crucial in simplifying such fractions effectively.
datafiend
Messages
31
Reaction score
0
Hi all,
I'm having a problem simplifying this:

[1/(1+x+h) - 1/(1+x)] / h

How do you get the common denominators for the top 2 fractions?

Thanks
 
Mathematics news on Phys.org
Maybe try to find a common denominator for 1+x+h and 1+x.

$(1+x+h)\times(1+h) = x^2+x(2+h)+1+h$
 
Last edited:
In the numerator, I would use:

$$\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}$$
 
mark,
thanks again!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top