Add three on both sides.

\(\displaystyle \sqrt{3x-3}-3+3 = 3+3\)

The equation becomes:  

\(\displaystyle \sqrt{3x-3} =6\)

Square both sides to eliminate the radical.

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

\(\displaystyle 3x-3 = 36\)

Add three on both sides.

\(\displaystyle 3x-3 +3= 36+3\)

\(\displaystyle 3x=39\)

Divide by three on both sides.

\(\displaystyle \frac{3x}{3}=\frac{39}{3}\)

The answer is:  \(\displaystyle 13\)

Divide by four on both sides.

\(\displaystyle \frac{4\sqrt[3]{-2x-3} }{4}= \frac{8}{4}\)

\(\displaystyle \sqrt[3]{-2x-3} = 2\)

Cube both sides to eliminate the radical.

\(\displaystyle (\sqrt[3]{-2x-3})^3 = 2^3\)

\(\displaystyle -2x-3=8\)

Add three on both sides.

\(\displaystyle -2x-3+3=8+3\)

\(\displaystyle -2x=11\)

Divide by negative two on both sides.

\(\displaystyle \frac{-2x}{-2}=\frac{11}{-2}\)

The answer is:  \(\displaystyle -\frac{11}{2}\)

Add 16 on both sides.

\(\displaystyle -\sqrt{x+8}-16+16= -20+16\)

\(\displaystyle -\sqrt{x+8} = -4\)

Divide both sides by negative one.

\(\displaystyle \frac{-\sqrt{x+8} }{-1}=\frac{ -4}{-1}\)

\(\displaystyle \sqrt{x+8}=4\)

Square both sides.

\(\displaystyle (\sqrt{x+8}) ^2=4^2\)

\(\displaystyle x+8=16\)

Subtract eight on both sides.

\(\displaystyle x+8-8=16-8\)

\(\displaystyle x=8\)

The answer is:  \(\displaystyle 8\)

Square both sides in order to eliminate the radical.

\(\displaystyle (\sqrt{-2x-5}) ^2= 9^2\)

\(\displaystyle -2x-5 = 81\)

Add 5 on both sides.

\(\displaystyle -2x-5 +5= 81+5\)

\(\displaystyle -2x=86\)

Divide by negative two on both sides.

\(\displaystyle \frac{-2x}{-2}=\frac{86}{-2}\)

Reduce both fractions.

The answer is:  \(\displaystyle -43\)

Add six on both sides.

\(\displaystyle \sqrt{3x}-6+6= 11+6\)

\(\displaystyle \sqrt{3x} = 17\)

Square both sides to eliminate the radical.

\(\displaystyle (\sqrt{3x}) ^2= 17^2\)

\(\displaystyle 3x=289\)

Divide both sides by three.

\(\displaystyle \frac{3x}{3}=\frac{289}{3}\)

The answer is:  \(\displaystyle \frac{289}{3}\)

Square both sides of the equation to eliminate the radical.

\(\displaystyle (\sqrt{2-3x})^2 = 40^2\)

The equation becomes:  

\(\displaystyle 2-3x=1600\)

Subtract two from both sides.

\(\displaystyle 2-3x-2=1600-2\)

\(\displaystyle -3x= 1598\)

Divide by negative three on both sides.

\(\displaystyle \frac{-3x}{-3}=\frac{ 1598}{-3}\)

The answer is:  \(\displaystyle -\frac{1598}{3}\)

Multiply by half on both sides.

\(\displaystyle 2\sqrt[3]{-2x} \cdot \frac{1}{2}= \frac{1}{2}\cdot \frac{1}{2}\)

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

Cube both sides to eliminate the radical.

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

\(\displaystyle -2x = \frac{1}{64}\)

Divide both sides by negative two.  This is the same as multiplying by negative half on the right side.

\(\displaystyle \frac{-2x}{-2} = \frac{1}{64}\cdot -\frac{1}{2}\)

Reduce both fractions.

The answer is:  \(\displaystyle -\frac{1}{128}\)

Multiply both sides by the denominator to eliminate the fraction on the left side.

\(\displaystyle \frac{x}{\sqrt{x-9}}\cdot \sqrt{x-9}= 6\cdot \sqrt{x-9}\)

The equation becomes: 

\(\displaystyle x= 6\sqrt{x-9}\)

Square both sides to eliminate the radical.

\(\displaystyle x^2=( 6\sqrt{x-9})^2\)

\(\displaystyle x^2 = 36(x-9)\)

Distribute the 36 into the binomial.

\(\displaystyle x^2 = 36x-324\)

Subtract \(\displaystyle 36x\) and add 324 on both sides of the equation.

\(\displaystyle x^2 -[36x]+(324)= 36x-324-[36x]+(324)\)

The equation becomes a parabola, which can be factored.

\(\displaystyle x^2-36x+324 = 0\)

\(\displaystyle (x-18)^2 = 0\)

The value of \(\displaystyle x=18\).  Check this value to see if substitution will satisfy the domain in the original equation.

\(\displaystyle \frac{18}{\sqrt{18-9}}\) is valid since there is no negative in the radical, and there is not a zero denominator.

The answer is:  \(\displaystyle 18\)

Subtract both sides by five.

\(\displaystyle \sqrt{-9x} +5-5=9-5\)

\(\displaystyle \sqrt{-9x} =4\)

Square both sides.

\(\displaystyle (\sqrt{-9x}) ^2=4^2\)

\(\displaystyle -9x = 16\)

Divide by negative nine on both sides.

\(\displaystyle \frac{-9x}{-9} = \frac{16}{-9}\)

The answer is:  \(\displaystyle -\frac{16}{9}\)

Add the radical on both sides.

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

The equation becomes:  

\(\displaystyle 2= 1+\sqrt[3]{4x}\)

Subtract 1 on both sides.

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

\(\displaystyle 1=\sqrt[3]{4x}\)

Cube both sides to eliminate the radical.

\(\displaystyle 1^3=(\sqrt[3]{4x})^3\)

\(\displaystyle 1=4x\)

Divide by four on both sides.

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

The answer is:  \(\displaystyle \frac{1}{4}\)