Now we can rewrite it as such

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

Now we need to get a common denominator.

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

Now we set up an equation to figure out \(\displaystyle A\) and \(\displaystyle B\).

\(\displaystyle x+10=A(x+3)+B(x-1)\)

To solve for \(\displaystyle A\), we are going to set \(\displaystyle x=1\).

\(\displaystyle 1+10=A(1+3)+B(1-1)\)

\(\displaystyle 11=4A\rightarrow A=\frac{11}{4}\)

To find \(\displaystyle B\), we need to set \(\displaystyle x=-3\)

\(\displaystyle -3+10=A(-3+3)+B(-3-1)\)

\(\displaystyle 7=-4B\rightarrow B=-\frac{7}{4}\)

Thus the answer is:

\(\displaystyle \frac{11}{4(x-1)}-\frac{7}{4(x+3)}\)

To add rational expressions, you must find the common denominator. In this case, it's \(\displaystyle x(x-1)\).

Next, you must change the numerators to offset the new denominator.

\(\displaystyle \frac{1}{x}\) becomes \(\displaystyle \frac{x-1}{x(x-1)}\)and \(\displaystyle \frac{4}{x-1}\) becomes \(\displaystyle \frac{4(x)}{(x-1)x}\).

Now you can combine the numerators: \(\displaystyle x-1+4x=5x-1\).

Put that over the denomiator and see if you can simplify/factor further. In this case, you can't.

Therefore, your final answer is:

\(\displaystyle \frac{5x-1}{x^2-1}\).

To subtract rational expressions, you must first find the common denominator, which in this case is \(\displaystyle x^2\). That means we only have to adjust the first fraction since the second fraction has that denominator already.

Therefore, the first fraction now looks like:

\(\displaystyle \frac{(x-1)(x)}{x(x)}\).

Now that the denominators are the same, combine numerators:

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

Now, put that over the denominator and see if you can simplify any further.

In this case, you can't, so your final answer is:
\(\displaystyle \frac{x^2-x-4}{x^2}\).

In order to add the numerators of the fractions, we need to find the least common denominator.

The least common denominator is:  \(\displaystyle ab\)

We will need to multiply the numerator and denominator by \(\displaystyle b\) to match the denominators of both fractions.

\(\displaystyle \frac{3}{ab} + \frac{b\left (a+3 \right )}{ab}\)

Simplify the fraction.

\(\displaystyle \frac{3}{ab} + \frac{b\left (a+3 \right )}{ab} = \frac{3}{ab}+\frac{ab+3b}{ab}\)

Combine the two fractions.

The answer is:  \(\displaystyle \frac{3+ab+3b}{ab}\)

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

The rules for adding fractions containing unknowns \(\displaystyle x\) are the same as for fractions containing explicit numbers, so you can guide yourself by recalling how you would proceed adding fractions such as,  

\(\displaystyle \frac{1}{3}+\frac{1}{5}\)

As you know you need to write them with a common denominator. In this case the least common denominator is \(\displaystyle 15\). So simply multiply the numerator and denominator of each fraction by the denominator of the other fraciton.

\(\displaystyle \frac{5}{5}\times\left(\frac{1}{3}\right)+\frac{3}{3}\times\left(\frac{1}{5}\right)\)

Notice that \(\displaystyle \frac{5}{5}\) and \(\displaystyle \frac{3}{3}\) are equal to one, this ensures that we are not changing the value of the fractions, we are changing only the representation of the value. 

\(\displaystyle = \frac{5}{15}+\frac{3}{15}\)

\(\displaystyle =\frac{8}{15}\)

Similairily, the procedure for an algebraic expression containing unknowns parallels this idea, 

\(\displaystyle \frac{x+3}{x+3}\times\left( \frac{1}{x}\right)+\frac{x}{x}\times\left(\frac{1}{x+3}\right)\)

Now we can add the numerators directly since we now have both terms expressed with a common denominator, \(\displaystyle x(x+3)\).

\(\displaystyle =\frac{x+3}{x(x+3)}+\frac{x}{x(x+3)}\)

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

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

1) First factor the denominator as much as possible; characterize the denominator and write the appropriate expansion:

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

 The denominator is a product of linear terms, so the partial fraction expansion will have the form, 

\(\displaystyle =\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+2}\)                                        (1.b)

2) Write a system of 3-equations and 3-unknowns in order to determine A,B,and C in the partial fraction expansion (equation 1.b).

If we were to take equation (1.b) and add each fraction under a common denominator \(\displaystyle x(x-3)(x+2)\), the numerator would have the form, 

\(\displaystyle A(x-3)(x+2)+Bx(x+2)+Cx(x-3)\)                (2.a)

Distribute and multiply the \(\displaystyle x\)'s,   

\(\displaystyle =A(x^2-x-6)+B(x^2+2x)+C(x^2-3x)\)           (2.b)

3) Find the constants A, B, and C. 

To find \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) simply expand and collect like terms (there are \(\displaystyle x^2\), \(\displaystyle x\), and constant terms) then compare to the original numerator \(\displaystyle 2x-1\).

 For the terms with \(\displaystyle x^2\) we must have, 

 \(\displaystyle Ax^2+Bx^2 +Cx^2 =(A+B+C)x^2= 0\)

So for \(\displaystyle x \neq 0\) we have,

 \(\displaystyle A+B+C = 0\)                                                                          (3.a)

For the \(\displaystyle x\)-terms we have, 

\(\displaystyle -Ax+2Bx-3Cx =(-A+2B-2C)x= 2x\)

for \(\displaystyle x \neq 0\) we have, 

\(\displaystyle -A+2B-3C=2\)                                                                 (3.b)

For the constant term we have, 

\(\displaystyle -6A = -1\)                                                                                  (3.c)

Right away we can read off the solution for \(\displaystyle A\) from equation (3.c) Substitute  \(\displaystyle A=\frac{1}{6}\)  into (3.a) and (3.b)  

4) Solve for the remaining unknown constants B and C, 

The system:

\(\displaystyle \frac{1}{6}+B+C = 0\)                                                                        (4.a)

\(\displaystyle -\frac{1}{6}+2B-3C=2\)                                                               (4.b)

In order to remove the fraction it would be convenient to solve this after multiplying both equations by \(\displaystyle 6\): 

\(\displaystyle 1+6B+6C=0\)                                                                   (4.c)

\(\displaystyle -1 +12B-18C=12\)                                                        (4.d)

 In order to make even more simple, multiply equation (4.c) by \(\displaystyle 2\) and solve for \(\displaystyle 12B\) in terms of \(\displaystyle C\) as follows,  

 \(\displaystyle 2+12B+12C = 0\)

\(\displaystyle 12B=-2-12C\)

Substitute into (4.d),

\(\displaystyle -1+(-2-12C)-18C=12\)

\(\displaystyle -30C=15\)

\(\displaystyle C = -\frac{1}{2}\)

Now we can use this value for \(\displaystyle C\) to find that \(\displaystyle B = \frac{1}{3}\). 

Finally, plug in the values for \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) we obtained into equation (1.b).

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

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

Factor the denominator: \(\displaystyle x^{2}+3x-18=(x+6)(x-3)\)

\(\displaystyle \frac{8x-42}{x^{2}+3x-18}=\frac{A}{x+6}+\frac{B}{x-3}\)

Multiply both sides of the equation by \(\displaystyle (x+6)(x-3)\)

\(\displaystyle 8x-42=A(x-3)+B(x+6)\)

Let \(\displaystyle x=3\): 

\(\displaystyle 24-42=B(3+6)\)

\(\displaystyle B=-2\)

Let \(\displaystyle x=-6\):

\(\displaystyle -48-42=A(-6-3)\)

\(\displaystyle A=10\)

\(\displaystyle \frac{8x-42}{x^{2}+3x-18}=\frac{10}{x+6}+\frac{-2}{x-3}\)

Factor the denominator: \(\displaystyle x^{2}-3x+2=(x-2)(x-1)\)

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

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

Multiply both sides of the equation by \(\displaystyle (x-2)(x-1)\)

\(\displaystyle 1=A(x-1)+B(x-2)\)

Let \(\displaystyle x=1\):

\(\displaystyle 1=B(1-2)\)

\(\displaystyle B=-1\)

Let \(\displaystyle x=2\):

\(\displaystyle 1=A(2-1)\)

\(\displaystyle A=1\)

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

Take the expression \(\displaystyle \frac{x+4}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}\) and multiply both sides by \(\displaystyle (x-1)(x+2)\). This transforms the equation into the following:

\(\displaystyle x+4=A(x+2)+B(x-1)\)

To solve for A we want to get rid of the B. So we set x equal to 1:

\(\displaystyle 1+4 = A(1+3)+B(1-1)\rightarrow 5 = 3A \rightarrow A=5/3\)

Similarly, we can get rid of A by setting x equal to -2:

\(\displaystyle -2+4=A(-2+2)+B(-2-1)\rightarrow2=-3B\rightarrow B=-2/3\)

Begin by distributing the exponent through the parentheses. The power rule dictates that an exponent raised to another exponent means that the two exponents are multiplied:

\(\displaystyle (3x^{4}y^2)(4xy^2)^{-3}= (3xy^2)(4^{-3}x^{-3}y^{-6})\)

Any negative exponents can be converted to positive exponents in the denominator of a fraction:

\(\displaystyle \frac{3x^4y^2}{64x^3y^6}\)

The like terms can be simplified by subtracting the power of the denominator from the power of the numerator:

\(\displaystyle \frac{3x^4y^2}{64x^3y^6}=\frac{3}{64}x^{4-3}y^{2-6}=\frac{3}{64}xy^{-4}\)

\(\displaystyle \frac{3x}{64y^4}\)