Question

❊ 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{ (1)/((a - b)(b - c)) + (1)/((c - b)(a - c) ) \: \: \{ANS : (1)/((a - b)(a - c) ) \}}}`

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Answer 1

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Answer 2

See Below.

Explanation:

Problem 1)

We want to simplify:

`\displaystyle (a+2)/(a^2+a-2)+(3)/(a^2-1)`

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

`\displaystyle =(a+2)/((a+2)(a-1))+(3)/((a+1)(a-1))`

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:

`\displaystyle =((a+2)(a+1))/((a+2)(a-1)(a+1))+(3(a+2))/((a+2)(a-1)(a+1))`

Add the fractions:

`\displaystyle =((a+2)(a+1)+3(a+2))/((a+2)(a-1)(a+1))`

Factor:

`\displaystyle =((a+2)((a+1)+3))/((a+2)(a-1)(a+1))`

Simplify:

`\displaystyle =(a+4)/((a-1)(a+1))`

We can expand. Therefore:

`\displaystyle =(a+4)/(a^2-1)`

Problem 2)

We want to simplify:

`\displaystyle (1)/((a-b)(b-c))+(1)/((c-b)(a-c))`

Again, let's create a common denominator. First, let's factor out a negative from the second term:

`\displaystyle \begin{aligned} \displaystyle &= (1)/((a-b)(b-c))+(1)/((-(b-c))(a-c))\\\\&=\displaystyle (1)/((a-b)(b-c))-(1)/((b-c)(a-c))\\\end{aligned}`

Now to create a common denominator, we can multiply the first term by (a - c) and the second term by (a - b). Hence:

`\displaystyle =((a-c))/((a-b)(b-c)(a-c))-((a-b))/((a-b)(b-c)(a-c))`

Subtract the fractions:

`\displaystyle =((a-c)-(a-b))/((a-b)(b-c)(a-c))`

Distribute and simplify:

`\displaystyle =(a-c-a+b)/((a-b)(b-c)(a-c))=(b-c)/((a-b)(b-c)(a-c))`

Cancel. Hence:

`\displaystyle =(1)/((a-b)(a-c))`