❊ Simplify : 1. [tex] \large{ \tt{ \frac{a + 2}{ {a}^{2} + a – 2 } + \frac{3}{ {a}^{2} – 1 } \: \: \{ANS : \frac{a + 4}{ {a}^{2} – 1 }

❊ Simplify :
1. \large{ \tt{ \frac{a + 2}{ {a}^{2} + a – 2 } + \frac{3}{ {a}^{2} – 1 } \: \: \{ANS : \frac{a + 4}{ {a}^{2} – 1 } \}}}

2. \large{ \tt{ \frac{1}{(a – b)(b – c)} + \frac{1}{(c – b)(a – c) } \: \: \{ANS : \frac{1}{(a – b)(a – c) } \}}}

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  1. 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 (ac) and the second term by (ab). 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]

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