In order to simplify this expression, first let's put the denominator of the first fraction in the factored form:
![ax^2+bx+c=a(x-x_1)(x-x_2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/dssedkcg4c691kz7ff1pn69fdovn9pno27.png)
To do so, let's find the zeros of the polynomial using the quadratic formula:
![\begin{gathered} x^2+12x+35=0 \\ a=1,b=12,c=35 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{-12+\sqrt[]{144-140}}{2}=(-12+2)/(2)=-5 \\ x_2=(-12-2)/(2)=-7 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/e53255iqjo5kyklnhco82fqsr8jxaur8fv.png)
So we have:
![x^2+12x+35=(x+5)(x+7)](https://img.qammunity.org/2023/formulas/mathematics/high-school/tl5v6yf5t8da6zj3x2ppwagrb2yb7vdog6.png)
Now, let's simplify the expression by inverting the division (turning it into a product) and canceling the like terms:
![\begin{gathered} ((x+7))/((x+5)(x+7))\colon((x+3))/((x+5)) \\ =(1)/(x+5)\cdot(x+5)/(x+3) \\ =(1)/(x+3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/4c11ih9rgq6ege084sfggbndo5km37r8m7.png)
The right side of the expression is correct, therefore the answer is TRUE.