In order to simplify this expression, we will need to multiply the numerator and denominator with the conjugate of the denominator.

$\displaystyle \frac{6i}{2i-1}\cdot \frac{2i+1}{2i+1}$

Simplify the top and bottom.  The value of $\displaystyle i=\sqrt{-1}$.

$\displaystyle 6i(2i+1) = 12i^2+6i = 12(-1)+6i = 6i-12$

$\displaystyle (2i-1)(2i+1) = 4i^2-1 = 4(-1)-1=-5$

Divide the numerator with the denominator.

$\displaystyle \frac{6i-12}{-5} = -\frac{6}{5}i+\frac{12}{5}$

The answer is:  $\displaystyle -\frac{6}{5}i+\frac{12}{5}$

Multiply the numerator and denominator by the conjugate of the denominator.

$\displaystyle \frac{i^3}{2i-4}\cdot \frac{2i+4}{2i+4} = \frac{2i^4+4i^3}{4i^2-16}$

Simplify the top and the bottom.

Recall that $\displaystyle i=\sqrt{-1}$, and $\displaystyle i^2= i\cdot i = -1$.

This means that:

$\displaystyle i^3 = i^2\cdot i = -i$

$\displaystyle i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

Re-substitute the actual values back into the fraction.

$\displaystyle \frac{2(1)+4(-i)}{4(-1)-16} = \frac{2-4i}{-20}$

Reduce and split this fraction.

The correct answer is:  $\displaystyle \frac{1}{5}i-\frac{1}{10}$

The values are imaginary.  Recall that:

$\displaystyle i=\sqrt{-1}$

This means that:  $\displaystyle i^2= i\cdot i = -1$

Rewrite the numerator.

$\displaystyle \frac{i^2+2i+1}{i-2} = \frac{-1+2i+1}{i-2}=\frac{2i}{i-2}$

Multiply the top and bottom by the conjugate of $\displaystyle i-2$.

$\displaystyle \frac{2i}{i-2}\cdot \frac{i+2}{i+2} = \frac{2i^2+4i}{i^2-4}$

Rewrite the terms using the value of $\displaystyle i^2$.

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

The answer is:  $\displaystyle -\frac{4}{5}i+\frac{2}{5}$

In order to simplify this, we will need to multiply the top and bottom of the fraction by the conjugate of the denominator.

$\displaystyle \frac{i-6}{2i-1} \times \frac{2i+1}{2i+1}$

Multiply the numerator.

$\displaystyle (i-6)(2i+1) = (i)(2i)+(i)(1)+(-6)(2i)+(-6)(1)$

$\displaystyle 2i^2+i-12i-6 = 2i^2-11i-6$

The term $\displaystyle i^2=-1$ since $\displaystyle i=\sqrt{-1}$.  Replace the term and simplify.

$\displaystyle = 2(-1)-11i-6 = -11i-8$

Multiply the denominator.

$\displaystyle (2i-1)(2i+1) =4i^2-1 = 4(-1)-1 =-5$

The expression becomes:

$\displaystyle \frac{i-6}{2i-1} \times \frac{2i+1}{2i+1} = \frac{ -11i-8}{-5}$

Dividing the double negatives will result into a positive value.

The answer is:  $\displaystyle \frac{11}{5}i+\frac{8}{5}$

Simplify the denominator.  Recall that $\displaystyle i=\sqrt{-1}$, and $\displaystyle i^2= i\cdot i = -1$.

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

Multiply the top and bottom by the conjugate of the denominator.  Use the FOIL method to simplify the denominator.

$\displaystyle \frac{i}{2i-2}\cdot \frac{2i+2}{2i+2} = \frac{2i^2+2i}{4i^2-4}$

Replace the values of the known imaginary terms.

$\displaystyle \frac{2i^2+2i}{4i^2-4} =\frac{2(-1)+2i}{4(-1)-4}=\frac{2i-2}{-8}$

Simplify this fraction.

The answer is:  $\displaystyle -\frac{1}{4}i+\frac{1}{4}$

In order to simplify this, we will need to multiply both the top and bottom by the conjugate of the denominator.

$\displaystyle \frac{10}{10+i} = \frac{10}{i+10}$

$\displaystyle \frac{10}{i+10} \times \frac{i-10}{i-10}$

Simplify the top and bottom.  The bottom can be simplified by the FOIL method.

$\displaystyle \frac{10}{i+10} \times \frac{i-10}{i-10} = \frac{10i-100}{i^2-100}$

The value of $\displaystyle i^2= i\cdot i = -1$ .  Replace the term.

$\displaystyle \frac{10i-100}{-1-100} = \frac{10i-100}{-101}$

Reduce this fraction.

The answer is:  $\displaystyle -\frac{10}{101}i+\frac{100}{101}$

In order to simplify this expression, we will need to multiply the numerator and denominator by the conjugate of the denominator.

$\displaystyle \frac{3i}{i+9}\cdot \frac{i-9}{i-9}$

Simplify both the top and bottom.  Recall that:  $\displaystyle i=\sqrt{-1}$

$\displaystyle 3i(i-9) = 3i^2-27i = 3(-1)-27i = -27i-3$

$\displaystyle (i+9)(i-9) = i^2-81 = (-1)-81 =-82$

Divide the numerator with the denominator.

$\displaystyle \frac{ -27i-3}{-82}$

Simplify this term and split the fraction.

The answer is:  $\displaystyle \frac{27}{82}i+\frac{3}{82}$

Multiply the top and bottom of the expression by the conjugate of the denominator.

$\displaystyle \frac{6i+6}{i-3}\times\frac{i+3}{i+3}$

Simplify the top and bottom using the FOIL method.

$\displaystyle (6i+6)(i+3) = 6i^2+18i+6i+18 = 6i^2+24i+18$

Note that:  $\displaystyle i^2= i\cdot i = -1$

Replace the term.  The numerator becomes:

$\displaystyle 6i^2+24i+18 =6(-1)+24i+18 = 24i+12$

Simplify the denominator.

$\displaystyle (i-3)(i+3)= i^2-9 = (-1)-9 = -10$

Divide the numerator with the denominator.

$\displaystyle \frac{ 24i+12}{-10} = -\frac{12}{5}i-\frac{6}{5}$

The answer is:  $\displaystyle -\frac{12}{5}i-\frac{6}{5}$

Multiply the top and bottom of this fraction by the conjugate of the denominator.

$\displaystyle \frac{3i}{3i-1}\cdot \frac{3i+1}{3i+1}$

Simplify the numerator.  Recall that $\displaystyle i=\sqrt{-1}$, and $\displaystyle i^2= i\cdot i = -1$.

$\displaystyle 3i(3i+1) = 9i^2+3i = 9(-1)+3i= 3i-9$

Simplify the denominator by FOIL method.

$\displaystyle (3i-1)(3i+1) = 9i^2-1 = 9(-1)-1=-10$

Divide the numerator with the denominator.

$\displaystyle \frac{3i-9}{-10}$

The answer is:  $\displaystyle -\frac{3}{10}i+\frac{9}{10}$

Multiply the top and the bottom of the fraction by the conjugate of the denominator.

$\displaystyle \frac{5}{i-7} \cdot \frac{i+7}{i+7}$

Expand the top and bottom.

$\displaystyle 5(i+7) = 5i+35$

$\displaystyle (i-7)(i+7) = i^2-49$

Recall that:

$\displaystyle i=\sqrt{-1}$

 $\displaystyle i^2= i\cdot i = -1$

Replace the imaginary term for $\displaystyle i^2$.

$\displaystyle \frac{5}{i-7} \cdot \frac{i+7}{i+7} = \frac{5i+35}{i^2-49} = \frac{5i+35}{(-1)-49}=\frac{5i+35}{-50}$

Simplify this fraction.

The answer is:  $\displaystyle -\frac{1}{10}i-\frac{7}{10}$