Question

Match each pairs of rational

Answer 1

Answer:

Refer to step-by-step.

Step-by-step explanation:

To find the LCD of the different fractions, we first need to follow the steps of finding the LCD of algebraic fractions.

1st step: Factor

You need to factor out the expressions if they do not have a difference of two squares.

2nd step: Find the LCD of the coefficients.

You need to find the LCD of the coefficients before moving onto the variables.

3rd step: The variable part of the LCD will be every single variable you see, with the highest exponent.

So let's begin, since all the given denominators are already factored out, we can proceed straight to step 2.

$\dfrac{5}{9m^{2}n}and\dfrac{5}{16mn^{3}}$

Now since we're dealing with the denominators, let's focus on them alone.

The LCD of 9 and 16 is 144

The LCD of variables m² and m is m²

The LCD of variables n and n³ is n²

So the LCD of the algebraic fractions $\dfrac{5}{9m^{2}n}and\dfrac{5}{16mn^{3}}$ is:

144m²n³

Now to the next one.

$\dfrac{5+m}{8m^{4} n^{2} }and\dfrac{5+n}{18m^{2} n^{4} }$

The LCD of 8 and 18 is 72

The LCD of variables $m^{4}$ and m² is $m^{4}$

The LCD of variable n² and $n^{4}$ is $n^{4}$

So the LCD of the algebraic fractions $\dfrac{5+m}{8m^{4} n^{2} }and\dfrac{5+n}{18m^{2} n^{4} }$is:

$72m^{4} n^{4}$

Next one we have:

$\dfrac{m+n}{12mn}and\dfrac{n+m}{18m^{3}n }$

The LCD of 12 and 18 is 36

The LCD of m and m³ is m³

The LCD of n and n is n

So the LCD of the algebraic fractions $\dfrac{m+n}{12mn}and\dfrac{n+m}{18m^{3}n }$ is:

36m³n

Last but not the least.

$\dfrac{m-n}{24mn^{4}}and\dfrac{m-n}{16m^{4}n }$

The LCD of 24 and 16 is 48

The LCD of m and $m^{4}$ is $m^{4}$

The LCD of $n^{4}$ and n is $n^{4}$

So the LCD of the algebraic fractions $\dfrac{m-n}{24mn^{4}}and\dfrac{m-n}{16m^{4}n }$ is:

$48m^{4} n^{4}$