Answer:
is the numerator.
Explanation:
The given term is:
![(x)/(x^(2)+3x+2 )+(3)/(x+1)](https://img.qammunity.org/2021/formulas/mathematics/college/4ka89j2c8koyia0x0h6ifq7rohgqzp1rjx.png)
First of all, let us have a look at the denominator of the 1st term:
![x^(2)+3x+2](https://img.qammunity.org/2021/formulas/mathematics/college/vwci08x5z8r2h89e3ul2m423oxshh3ykwm.png)
Factorizing by writing
as
and then taking 'x' and '1' common respectively:
![x^(2)+2x+x+2 \\\Rightarrow x(x+2)+1(x+2 )\\\Rightarrow (x+1)(x+2 )](https://img.qammunity.org/2021/formulas/mathematics/college/khl81cqor1ycqigmbdtc31ix84hjln71xp.png)
Now, solving the given expression by taking LCM:
![\frac{x}{(x+{2})(x+1)}+(3)/(x+1)\\\Rightarrow \frac{x+3(x+2)}{(x+{2})(x+1)}\\\Rightarrow \frac{x+3x+6}{(x+{2})(x+1)}\\\Rightarrow \frac{4x+6}{(x+{2})(x+1)}](https://img.qammunity.org/2021/formulas/mathematics/college/ceae6rwngq6jv3bdxdwx4c2c7gmily8awe.png)
Any expression
has
as its numerator and
as its denominator.
So, the numerator of simplified term is:
![4x+6](https://img.qammunity.org/2021/formulas/mathematics/college/3ksg6cmpahx39vuvafrvy4l1bpmhojhvv0.png)