Remember that that fraction bar is just a division symbol. Rewrite as a division, rewrite those mixed fractions as improper fractions, then rewrite as as a multiplication by the reciprocal of the second fraction.

\(\displaystyle \frac{2\tfrac{1}{4}}{5\tfrac{2}{5}} = 2\tfrac{1}{4} \div 5\tfrac{2}{5} = \frac{9}{4} \div \frac{27}{5} = \frac{9}{4} \cdot \frac{5}{27} = \frac{1}{4} \cdot \frac{5}{3}= \frac{5}{12}\)

Find the least common denominator which is \(\displaystyle x^2\).

Just multiply the right fraction top and bottom by \(\displaystyle x\). 

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

Finally, subtract. 

Answer should be, 

 \(\displaystyle \frac{2-4x}{x^2}\).

Lets try to factor. Remember, we need to find two terms that are factors of the c term that add up to the b term. 

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

Next cancel out the like terms.

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

Now combine the numerator. Remember to distribute the negative sign.

This is the final answer:

 \(\displaystyle \frac{-1}{x+4}\).

Turn the \(\displaystyle 1\) into a fraction that has a common denominator with the other fraction. To do this multiply \(\displaystyle 1 \cdot \frac{x-3}{x-3}\).

This results in the following expression:

\(\displaystyle \frac{x-3}{x-3}-\frac{3}{x-3}\)

With the same denominator, just subtract and remember to distribute the negative sign.

The final answer is

 \(\displaystyle \frac{x-6}{x-3}\).

Lets try to reduce the fraction. When factoring a difference of squares, the form is \(\displaystyle (x-a)(x+a)\). When you foil, the middle terms cancel out.

So when doing that, we have:

\(\displaystyle \frac{(x-3)(x+3)}{(x+3)}-\frac{(x-4)(x+4)}{(x+4)}\).

Cancel out like terms and we get this:

\(\displaystyle (x-3)-(x-4)\).

Distribute the negative sign and we should get \(\displaystyle 1\).

First find the least common denominator. That will be  \(\displaystyle 2\sqrt{x}\) .

Multiply top and bottom of the left fraction by \(\displaystyle 2\) .

\(\displaystyle \frac{1}{\sqrt{x}}\cdot \frac{2}{2}=\frac{2}{2\sqrt{x}}\)

Then subtract the numerator and then cross-multiply to get this: 

\(\displaystyle \frac{2}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=2\rightarrow 2-1=2(2\sqrt{x}) \rightarrow 1=4\sqrt{x}\)

Then combine like terms. To get rid of the square root,

\(\displaystyle \frac{1}{4}=\sqrt{x}\) square both sides to get \(\displaystyle \left(\frac{1}{4}\right)^2=\frac{1}{16}\) as the final answer.

Find the least common denominator which is \(\displaystyle x^2-4\). Then multiply the left fraction numerator by \(\displaystyle x+2\) and multiply the right numerator by \(\displaystyle x-2\) inorder for each fraction to share the common denominator.

\(\displaystyle \frac{5(x+2)}{x^2-4}-\frac{2(x-2)}{x^2-4}=1\) 

Distribute and be careful of the negative sign.

\(\displaystyle \frac{5x+10-2x+4}{x^2-4}=1\) 

Cross-multiply and create the quadratic equation.

\(\displaystyle x^2-3x-18=0\) 

Lets factor. Remember, we need to find two terms that are factors of the c term that add up to the b term. 

\(\displaystyle (x-6)(x+3)=0\)

\(\displaystyle x=6\), \(\displaystyle x=-3\)

If you check these answers back into the question, none of the fractions are undefined so these are the final answers. 

Find the least common denominator which is \(\displaystyle x^2-1\). Then multiply the left fraction numerator by \(\displaystyle x+1\) and multiply the right numerator by \(\displaystyle x-1\) in order to make each fraction share the common denominator.

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

Distribute and be careful of the negative sign.

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

Simplify the numerator and cross multiply. Because there is an absolute value bar, we need to split this expression into two different equations.

Equation one:

\(\displaystyle 2(x^2-1)=4\)

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

\(\displaystyle 2x^2=6\)

\(\displaystyle x^2=3\)

\(\displaystyle x=\pm \sqrt{3}\) By inspection, these values pass and don't violate the fractions being undefined.

Equation two:

\(\displaystyle -2(x^2-1)=4\)

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

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

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

\(\displaystyle x=i\) This answer is imaginary and not in the choices leaving \(\displaystyle \pm \sqrt{3}\) as the answers.

Keep in mind that the main fraction line just means "divide." 

So then,

\(\displaystyle \frac{\frac{2}{3}}{\frac{4}{5}}=\frac{2}{3}\div\frac{4}{5}\)

Now, divide this like any other fraction.

\(\displaystyle \frac{2}{3}\div\frac{4}{5}=\frac{2}{3}\times\frac{5}{4}=\frac{10}{12}=\frac{5}{6}\)

Remember that the main line in the fraction means "divide."

Then,

\(\displaystyle \frac{\frac{4}{3}}{6}=\frac{4}{3}\div6\)

Now, divide this like you would divide any other fraction.

\(\displaystyle \frac{4}{3}\div6=\frac{4}{3}\times\frac{1}{6}=\frac{4}{18}=\frac{2}{9}\)