Notice that the second fraction can be simplified.  Factorize the denominator.

\(\displaystyle x^2+4x-21 = (x-3)(x+7)\)

Rewrite the expression.

\(\displaystyle \frac{1}{x}-\frac{x-3}{x^2+4x-21}=\frac{1}{x}-\frac{x-3}{(x-3)(x+7)}\)

Reduce the fraction.

\(\displaystyle \frac{1}{x}-\frac{1}{x+7}\)

Multiply both denominators together to determine the least common denominator.

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

Convert the fractions and solve.

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

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

Determine the least common denominator.

\(\displaystyle 9x(x+1) = 9x^2+9x\)

Convert the fractions.

\(\displaystyle \frac{3x(9x)}{(x+1)(9x)} + \frac{7(x+1)}{(9x)(x+1)}\)

The expression becomes:

\(\displaystyle \frac{3x(9x)+7(x+1)}{9x^2+9x} = \frac{27x^2+7x+7}{9x^2+9x}\)

The answer is:  \(\displaystyle \frac{27x^2+7x+7}{9x^2+9x}\)

Determine the least common denominator.

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

Convert the fractions.

\(\displaystyle \frac{9}{x}-\frac{x}{x-1}=\frac{9(x-1)}{x^2-x}-\frac{x(x)}{x^2-x}\)

Simplify the numerator.

\(\displaystyle \frac{9(x-1)}{x^2-x}-\frac{x(x)}{x^2-x} = \frac{9x-9}{x^2-x}-\frac{x^2}{x^2-x}\)

Combine as one fraction.

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

Pull out a common factor of negative one.  This will allow us to pull the negative in front of the fraction.

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

The answer is:  \(\displaystyle -\frac{x^2-9x+9}{x^2-x}\)

In order to add both terms, we will need to find the least common denominator.

Multiply the denominators together.

\(\displaystyle x(3x+9) = 3x^2+9x\)

Convert the two fractions.

\(\displaystyle \frac{1}{3x+9}+\frac{2}{x}=\frac{x}{3x^2+9x}+\frac{2(3x+9)}{3x^2+9x} =\frac{7x+18}{3x^2+9x}\)

The answer is:  \(\displaystyle \frac{7x+18}{3x^2+9x}\)

In order to subtract the numerators, we will need to determine the least common denominator.  Upon visualization, the denominators share an x term. This means that we will not have to change the x term.

Multiply the first denominator by ten, and the second denominator by three.

\(\displaystyle \frac{1}{3x}-\frac{3}{10x}=\frac{1(10)}{3x(10)}-\frac{3(3)}{10x(3)}\)

Simplify the numerator.

\(\displaystyle \frac{1(10)}{3x(10)}-\frac{3(3)}{10x(3)} = \frac{10}{30x}- \frac{9}{30x}\)

The answer is:  \(\displaystyle \frac{1}{30x}\)

Multiply the denominators to get the least common denominator.  We can then convert both fractions so that the denominators are alike.

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

Simplify both the top and the bottom.

\(\displaystyle \frac{6x}{-2x^2+4x} - \frac{10-5x}{-2x^2+4x}\)

Combine the numerators as one fraction.  Be careful with the second fraction since the entire numerator is a quantity, which means we will need to brace \(\displaystyle 10-5x\) with parentheses.

\(\displaystyle \frac{6x-(10-5x)}{-2x^2+4x} =\frac{6x-10+5x}{-2x^2+4x} = \frac{11x-10}{-2x^2+4x}\)

Pull out a common factor of negative one in the denominator.  This allows us to rewrite the fraction with the negative sign in front of the fraction.

\(\displaystyle \frac{11x-10}{-2x^2+4x}= \frac{11x-10}{-(2x^2-4x)}= -\frac{11x-10}{2x^2-4x}\)

The answer is:  \(\displaystyle -\frac{11x-10}{2x^2-4x}\)

In order to solve this expression, we will need to determine the least common denominator.  Notice that the denominators both share an x term.  We do not need to change that.

Multiply the denominator of the first fraction by two to get the least common denominator, which is \(\displaystyle 10x\).

\(\displaystyle \frac{7(2)}{5x(2)}+\frac{9}{10x} = \frac{14}{10x}+\frac{9}{10x}\)

Add the numerators.

The answer is:  \(\displaystyle \frac{23}{10x}\)

Determine the least common denominator by multiplying the denominators together.

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

Convert the fractions given.

\(\displaystyle \frac{9x}{x+1}-\frac{9}{x}=\frac{(x)(9x)}{(x)(x+1)}-\frac{9(x+1)}{x(x+1)}\)

Simplify the numerators and denominators.

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

Combine the terms as one fraction.  Make sure to brace the \(\displaystyle 9x+9\) term since this is a quantity.

\(\displaystyle \frac{9x^2}{x^2+x}-\frac{9x+9}{x^2+x} = \frac{9x^2-(9x+9)}{x^2+x}\)

The answer is:  \(\displaystyle \frac{9x^2-9x-9}{x^2+x}\)

Identify the least common denominator by multiplying the denominators together.

Convert the fractions.

\(\displaystyle \frac{4x}{x-1}-\frac{10}{x} = \frac{4x(x)}{x(x-1)}-\frac{10(x-1)}{x(x-1)}\)

Simplify the numerator and denominator.

\(\displaystyle \frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}\)

Combine both fractions as one.  Make sure to enclose the second number in parentheses since the negative sign is distributive.

\(\displaystyle \frac{4x^2}{x^2-x}-\frac{10x-10}{x^2-x}=\frac{4x^2-(10x-10)}{x^2-x} =\frac{4x^2-10x+10}{x^2-x}\)

The answer is:  \(\displaystyle \frac{4x^2-10x+10}{x^2-x}\)

Evaluate by changing the denominator of the second fraction so that both fractions have similar denominators.

\(\displaystyle \frac{7}{9x}+\frac{2(4-x)}{x} = \frac{7}{9x}+\frac{2[9](4-x)}{[9]x}\)

Use distribution to simplify the numerator.

\(\displaystyle \frac{7}{9x}+\frac{2[9](4-x)}{[9]x} =\frac{7}{9x}+\frac{72-18x}{9x}\)

Combine like-terms.

The answer is:  \(\displaystyle \frac{-18x+79}{9x}\)