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mezya [45]
3 years ago
12

❊ Simplify :

Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

\displaystyle \frac{a+2}{a^2+a-2}+\frac{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 =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}

Now, create a common denominator. To do this, we can multiply the first term by (<em>a</em> + 1) and the second term by (<em>a</em> + 2). Hence:

\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}

Add the fractions:

\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}

Factor:

\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}

Simplify:

\displaystyle =\frac{a+4}{(a-1)(a+1)}

We can expand. Therefore:

\displaystyle =\frac{a+4}{a^2-1}

Problem 2)

We want to simplify:

\displaystyle \frac{1}{(a-b)(b-c)}+\frac{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 &= \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}

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

\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}

Subtract the fractions:

\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}

Distribute and simplify:

\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}

Cancel. Hence:

\displaystyle =\frac{1}{(a-b)(a-c)}

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