Answer:
See Below.
Step-by-step explanation:
Problem 1)
We want to simplify:
[tex]\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}[/tex]
First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:
[tex]\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}[/tex]
Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:
[tex]\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}[/tex]
Add the fractions:
[tex]\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}[/tex]
Factor:
[tex]\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}[/tex]
Simplify:
[tex]\displaystyle =\frac{a+4}{(a-1)(a+1)}[/tex]
We can expand. Therefore:
[tex]\displaystyle =\frac{a+4}{a^2-1}[/tex]
Problem 2)
We want to simplify:
[tex]\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}[/tex]
Again, let's create a common denominator. First, let's factor out a negative from the second term:
[tex]\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}[/tex]
Now to create a common denominator, we can multiply the first term by (a - c) and the second term by (a - b). Hence:
[tex]\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}[/tex]
Subtract the fractions:
[tex]\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}[/tex]
Distribute and simplify:
[tex]\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}[/tex]
Cancel. Hence:
[tex]\displaystyle =\frac{1}{(a-b)(a-c)}[/tex]
