Answer:
![((x+5)(x+2))/(x(x-3)(x+3))](https://img.qammunity.org/2021/formulas/mathematics/college/1dg6oxt5cnv984lvlb4z60iicxs3lqcp4b.png)
Explanation:
The given expression is
![(2x+5)/(x^(2) -3x) -(3x+5)/(x^(3) -9x) -(x+1)/(x^(2)-9)](https://img.qammunity.org/2021/formulas/mathematics/college/3r6rwhi9sb3vcuw2w6jzw0yli7itnaztqb.png)
First, we need to factor each denominator
![(2x+5)/(x(x-3)) -(3x+5)/(x(x+3)(x-3)) -(x+1)/((x-3)(x+3))](https://img.qammunity.org/2021/formulas/mathematics/college/l3390vfsl3f2svnuw2tvc318mt07ttrrtc.png)
So, the least common factor (LCF) is
, because they are the factors that repeats.
Now, we diviide the LCF by each denominator, to then multiply it by each numerator.
![((x+3)(2x+5)-3x-5-x(x+1))/(x(x-3)(x+3)) =(2x^(2)+5x+6x+15-3x-5-x^(2)-x )/(x(x-3)(x+3))\\(x^(2)+7x+10)/(x(x-3)(x+3))](https://img.qammunity.org/2021/formulas/mathematics/college/qht3ch8ar7uu3zledrp7f6zlaz0e9g8wqb.png)
Then, we factor the numerator, to do so, we need to find two numbers which product is 10 and which sum is 7.
![((x+5)(x+2))/(x(x-3)(x+3))](https://img.qammunity.org/2021/formulas/mathematics/college/1dg6oxt5cnv984lvlb4z60iicxs3lqcp4b.png)
Therefore, the expression is equivalent to
![((x+5)(x+2))/(x(x-3)(x+3))](https://img.qammunity.org/2021/formulas/mathematics/college/1dg6oxt5cnv984lvlb4z60iicxs3lqcp4b.png)