Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't get this one either for whatever reason, spent about 20 minutes on it and I have made no progress at all.

$$\frac{x^2}{(x^2-4)} - \frac{x+1}{x+2}.$$

I know that I can simplify this into one fraction so I make it $$\frac{x^2}{x+2}-\frac{(x+1)(x^2-4)}{(x^2-4)(x+2)}$$

I then can simplify it further making the $(x^2+4)$ into $(x-2)(x+2)$ and the $(x+2)$ into $(x-1)(x+1)$ but this does not help me get the answer. I know I have to manipulate it in some counter intuitive way but I can not make it work.

share|cite|improve this question
I edited the question; please check if it's still what you meant. – Davide Giraudo Dec 17 '11 at 23:13
Yes thank you, I have no idea why I could not get the fractions to work. – user138246 Dec 17 '11 at 23:15
Type '\frac{a+b}{c+d}' to get $\frac{a+b}{c+d}$. – Davide Giraudo Dec 17 '11 at 23:18
In the original expression, multiply top and bottom by $x-2$. That will make the denominator of that term $(x+2)(x-2)$, which is $x^2-4$, so you will get a common denominator. Or else, but this is more unpleasant, multiply top and bottom of first term by $x+2$, and top and bottom of second term by $x^2-4$. That will get you a common denominator of $(x+2)(x^2-4)$. – André Nicolas Dec 17 '11 at 23:18
up vote 3 down vote accepted

As first step we may use the common denominator $(x^{2}-4)=(x-2)(x+2)$ because the $\text{lcm}\left( (x-2)(x+2),(x+2)\right) =(x-2)(x+2)$ $$ \begin{eqnarray*} \frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2} &=&\frac{x^{2}}{(x-2)(x+2)}-\frac{ \left( x+1\right) (x-2)}{\left( x+2\right) (x-2)} \\ &=&\frac{x^{2}-\left( x+1\right) (x-2)}{(x-2)(x+2)}.\tag{1} \end{eqnarray*} $$

Otherwise we would get the equivalent but more more complex fraction $$ \frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2}=\frac{x^{2}\left( x+2\right) -\left( x+1\right) (x^{2}-4)}{(x^{2}-4)\left( x+2\right) }. $$ Expanding the second term of the numerator of $(1)$ $$ \begin{eqnarray*} \left( x+1\right) (x-2) &=&x(x-2)+(x-2)=x^{2}-2x+x-2 \\ &=&x^{2}-x-2 \end{eqnarray*} $$

and substituting into the fraction we get $$ \frac{x^{2}-\left( x^{2}-x-2\right) }{(x-2)(x+2)}=\frac{x^{2}-x^{2}+x+2}{ (x-2)(x+2)}=\frac{x+2}{(x-2)(x+2)},\tag{2} $$ which for $x+2\neq 0$ simplifies to $$ \frac{1}{x-2}\tag{3} $$

Added: In general we transform the sum (or difference) of two given rational fractions (the numerator and denominator consists of polynomials) into a single equivalent fraction, by using properties such as

  1. $$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$$
  2. $$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$
  3. $$\frac{A(x)}{B_1(x)B_2(x)}\pm \frac{C(x)}{B_2(x)}=\frac{A(x)\pm B_1(x)C(x)}{B_1(x)B_2(x)}.$$
share|cite|improve this answer

$$\frac{x^2}{x^2-4} - \frac{x+1}{x+2} = \frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x-2)(x+1)}{(x-2)(x+2)}.$$

Working with the numerators: $$ x^2-(x-2)(x+1) = x^2 - (x^2 -x -2). $$ Here's the easiest mistake to make (I've seen this happen zillions of times including in calculus courses):

Right: $x^2 - (x^2 -x -2) = x^2 - x^2 + x + 2$

Wrong: $x^2 - (x^2 -x -2) = x^2 - x^2 - x - 2$

The numerator ends up being $x+2$, so we get another simplification: $$ \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}. $$

share|cite|improve this answer

It definitely makes life easier to notice that $x^2 - 4 = (x + 2)(x - 2)$. If you multiply the second term of the original expression by \[ 1 = \frac{x - 2}{x - 2} \] then you get \[ \frac{x^2}{(x + 2)(x - 2)} - \frac{(x + 1)(x - 2)}{(x + 2)(x - 2)}. \] These two fractions have a common denominator, so you can combine them into a single quotient. Work out the numerator, and see if you can cancel anything after that.

Some comments on your attempt: It isn't true that $x + 2$ equals $(x + 1)(x - 1) = x^2 - 1$. It is a good idea to look for differences of squares, though. It's certainly fine (perhaps a bit messier) to place everything over the common denominator $(x^2 - 4)(x + 2)$, but for this you would multiply the first term of your original expression by $1 = (x + 2)/(x + 2)$; could you explain how you got $x^2/(x + 2)$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.