In order to solve this rational expression, denominators must be common.

Multiply both denominators together.

\(\displaystyle 3x(x-1) = 3x^2-3x\)

Convert the fractions with the same base denominator.  Multiply the numerator with what was multiplied on the denominator to achieve the new denominator.

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

Combine the two fractions into a single fraction.

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

Simplify the numerator on the right side.

\(\displaystyle \frac{3x+2(x-1)}{3x^2-3x} = \frac{3x+2x-2}{3x^2-3x}= \frac{5x-2}{3x^2-3x}\)

The answer is:  \(\displaystyle \frac{5x-2}{3x^2-3x}\)

Rewrite the fractions with the least common denominator.  The least common denominator is \(\displaystyle x^3\).  We will need both fractions to have this denominator in order to add or subtract the numerator.

\(\displaystyle \frac{(6x-1)(x^2)}{x(x^2)}-\frac{3+x}{x^3}\)

Distribute the numerator and denominator of the first term.

\(\displaystyle \frac{6x^3-x^2}{x^3}-\frac{3+x}{x^3}\)

Combine the numerators as one fraction.  Be sure to enclose the second term in parentheses.

\(\displaystyle \frac{6x^3-x^2-(3+x)}{x^3}\)

Simplify the numerator.

\(\displaystyle \frac{6x^3-x^2-3-x}{x^3}\)

There are no common factors.

The answer is:  \(\displaystyle \frac{6x^3-x^2-x-3}{x^3}\)

In order to add the numerators, we will need the least common denominator.

Multiply the denominators together.

\(\displaystyle x(x+7)=x^2+7x\)

Convert both fractions by multiplying both the top and bottom by what was multiplied to get the denominator.  Rewrite the fractions and combine as one single fraction.

\(\displaystyle \frac{5(x+7)}{x(x+7)}-\frac{2x(x)}{(x)(x+7)} = \frac{5x+35-2x^2}{x^2+7x}\)

Re-order the terms.

\(\displaystyle \frac{-2x^2+5x+35}{x^2+7x}\)

Pull out a common factor of negative one on the numerator.

\(\displaystyle \frac{-1(2x^2-5x-35)}{x^2+7x} = -\frac{2x^2-5x-35}{x^2+7x}\)

The answer is:  \(\displaystyle -\frac{2x^2-5x-35}{x^2+7x}\)

In order to subtract the fractions, multiply both denominators together in order to obtain the least common denominator.

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

Simplify the numerators.

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

Combine the numerators.

\(\displaystyle =\frac{2x^2-(2x+1)}{(2x+1)(2x)}=\frac{2x^2-2x-1}{4x^2+2x}\)

The answer is:  \(\displaystyle \frac{2x^2-2x-1}{4x^2+2x}\)

To simplify this expression, we will need to multiply both denominators together to find the least common denominator.

\(\displaystyle x(x+3) =x^2+3x\)

Convert both fractions to the common denominator.

\(\displaystyle \frac{2}{x+3}+\frac{8}{x}=\frac{2x}{x^2+3x}+\frac{8(x+3)}{x^2+3x}\)

\(\displaystyle 8(x+3)=8x+24\)

Combine the fractions.

\(\displaystyle \frac{2x+8x+24}{x^2+3x} = \frac{10x+24}{x^2+3x}\)

The answer is:  \(\displaystyle \frac{10x+24}{x^2+3x}\)

To add these rational expressions, first identify the common denominator.

In this case, it's the product of the two denominators:

\(\displaystyle (x+2)(x-2)=x^2-4\).

Then, multiply the numerators to offset them for their new denominator:

\(\displaystyle \frac{4(x-2)}{(x+2)(x-2)}+\frac{9x(x+2)}{(x-2)(x+2)}\).

Then combine the numerators to get

\(\displaystyle 4x-8+9x^2+18x=9x^2+22x-8\).

Thus, your final answer is:

\(\displaystyle \frac{9x^2+22x-8}{x^2-4}.\)

To add these two expressions, first identify the common denominator.

In this case, it's the product of the two, which is

\(\displaystyle x^2(x-1)\) .

Then, multiply the numerators to offset them for the new denominator:

\(\displaystyle \frac{4(x-1)}{x^2(x-1)}+\frac{9(x^2)}{x^2(x-1)}\).

Now, combine the numerators to get your answer of:

\(\displaystyle \frac{9x^2+4x-4}{x^2(x-1)}.\)

First, identify the common denominator. In this case, it's \(\displaystyle x^3\).

Offset the first two numerators to make up for the new denominator:

\(\displaystyle \frac{9(x^2)}{x^3}+\frac{2(x)}{x^3}-\frac{5}{x^3}\).

Now, combine the numerators to get:

\(\displaystyle 9x^2+2x-5\).

Thus, your final answer is:

\(\displaystyle \frac{9x^2+2x-5}{x^3}.\)

Find the least common denominator by multiplying the denominators together.

\(\displaystyle (1-x)(3+x) = (1)(3)+(1)(x)+(-x)(3)+(-x)(x)\)

Simplify the terms.

\(\displaystyle (1-x)(3+x)= -x^2-2x+3\)

Convert both fractions.

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

Simplify the numerator and denominator.

\(\displaystyle \frac{6+2x}{(1-x)(3+x)}-\frac{2-2x}{(3+x)(1-x)}\)

Combine to form one fraction.  Brace the numerator of the second term since we are subtracting a quantity.

\(\displaystyle \frac{6+2x-[2-2x]}{(1-x)(3+x)} =\frac{6+2x-2+2x}{(1-x)(3+x)} =\frac{4x+4}{(1-x)(3+x)}\)

Replace the denominator with the simplified form.

\(\displaystyle \frac{4x+4}{(1-x)(3+x)} =\frac{4x+4}{-x^2-2x+3}\)

Pull out a common factor of negative one from the denominator.

\(\displaystyle \frac{4x+4}{-x^2-2x+3} = \frac{4x+4}{-1(x^2+2x-3)}\)

This allows us to pull the negative sign out in front of the fraction.

The answer is:  \(\displaystyle -\frac{4x+4}{x^2+2x-3}\)

Determine the least common denominator by multiplying both denominator together.  Convert the fractions.

\(\displaystyle \frac{2}{x}+\frac{2}{5x^2+3}=\frac{2(5x^2+3)}{x(5x^2+3)}+\frac{2x}{x(5x^2+3)}\)

Combine the fractions as one fractions and simplify the numerator.

\(\displaystyle \frac{2(5x^2+3)+2x}{x(5x^2+3)}= \frac{10x^2+6+2x}{x(5x^2+3)}= \frac{10x^2+2x+6}{5x^3+3x}\)

The answer is:  \(\displaystyle \frac{10x^2+2x+6}{5x^3+3x}\)