In order to solve for \(\displaystyle \frac{5-2}{\sqrt{x}}=\frac{3}{4}\), cross multiplication must be used. Applying this, the equestion to be solved will be:

\(\displaystyle 4(5-2)=3\sqrt{x}\)

\(\displaystyle 4(5-2)=3\sqrt{x}\)

\(\displaystyle 4\cdot3=3\sqrt{x}\)

Next, each side is divided by 3. 

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

Next, each side is squared. 

\(\displaystyle x=16\)

\(\displaystyle x=\frac{125}{5x}\)

The first step to solve for x is to multiply each side by 5x. This results in:

\(\displaystyle 5x^{2}=125\)

Divide each side by 5. 

\(\displaystyle x^{2}=25\)

Next, take the square root of each side. This results in:

\(\displaystyle x=\pm 5\)

To start, use the distributive property on the left side: \(\displaystyle 2(x-5)=2x-10\), so that you then have \(\displaystyle 2x-10=3x+12\). Then, combine like terms to get \(\displaystyle 1x=-22\) so that your final answer is \(\displaystyle x=-22.\)

The absolute value of a nonnegative number is the number itself; the absolute value of a negative number is its positive opposite.

By substitution, 20 can be seen to be a solution of each of the equations in the four choices.

\(\displaystyle \left | x - 2 \right | = 18\)

\(\displaystyle \left | 20 - 2 \right | = 18\)

\(\displaystyle \left |18 \right | = 18\) - true.

20 can be confirmed as a solution to the other three equations similarly. Therefore, the question is essentially to choose the equation with \(\displaystyle -16\) as a solution. Substituting \(\displaystyle -16\) for \(\displaystyle x\) in each equation:

\(\displaystyle \left | x - 2 \right | = 18\)

\(\displaystyle \left | -16 - 2 \right | = 18\)

\(\displaystyle |-18| = 18\) - true. This is the correct choice.

As for the other three:

\(\displaystyle \left | -16 - 3 \right | = 17\)

\(\displaystyle |-19| = 17\)

\(\displaystyle 19 = 17\) - false. 

The other two equations can be similarly proved to not have \(\displaystyle -16\) as a solution.

\(\displaystyle \frac{1}{x} = 12\); taking the reciprocal of both sides, 

\(\displaystyle x = \frac{1}{\left (\frac{1}{x} \right )} = \frac{1}{12}\)

One tenth of \(\displaystyle x\) is 

\(\displaystyle \frac{1}{10} \cdot x = \frac{1}{10} \cdot \frac{1}{12} = \frac{1}{120}\)

Take the reciprocal of both sides of the equation, then solve:

\(\displaystyle \frac{1}{x+2} = 12\)

\(\displaystyle x+2 = \frac{1}{12}\)

\(\displaystyle x+2 - 2 = \frac{1}{12} - 2\)

\(\displaystyle x = \frac{1}{12} - \frac{24}{12} = \frac{1-24}{12} = -\frac{23}{12}\)

One sixth of this is 

\(\displaystyle \frac{1}{6} x = \frac{1}{6}\left ( -\frac{23}{12} \right ) = -\frac{23}{6 \cdot 12} = - \frac{23}{72}\)

\(\displaystyle \left | 2x- 7\right | + 49 = 20\)

\(\displaystyle \left | 2x- 7\right | + 49 - 49 = 20 - 49\)

\(\displaystyle \left | 2x- 7\right | = -29\)

Since it is impossible for the absolute value of a number to be negative, the equation has no solution.

\(\displaystyle 8x^{3} + 12x^{2} = 18 x + 27\)

\(\displaystyle 8x^{3} + 12x^{2} - 18 x - 27 = 18 x + 27 - 18 x - 27\)

\(\displaystyle 8x^{3} - 18 x + 12x^{2}- 27 = 0\)

\(\displaystyle (8x^{3} - 18 x )+ (12x^{2}- 27) = 0\)

\(\displaystyle 2x (4x^{2} - 9 )+ 3 (4x^{2}- 9) = 0\)

\(\displaystyle (2x + 3 )(4x^{2} - 9 ) = 0\)

\(\displaystyle (2x + 3 ) [ (2x)^{2} - 3^{2} ]= 0\)

\(\displaystyle (2x + 3 ) (2x + 3 ) (2x - 3 ) = 0\)

One of two things must happen - either

\(\displaystyle 2x+3 = 0\)

or 

\(\displaystyle 2x-3 = 0\)

This gives this equation two real solutions - the solutions to these two equations.

\(\displaystyle 12x^{2} + 26x = 56\)

\(\displaystyle 12x^{2} + 26x - 56 = 56 - 56\)

\(\displaystyle 12x^{2} + 26x - 56 = 0\)

\(\displaystyle 2 (6x^{2} + 13x - 28 ) = 0\)

The quadratic trinomial can be factored using the \(\displaystyle ac\) method by looking for two integers whose sum is 13 and whose product is \(\displaystyle 6 \times (-28 ) = -168\). Throught trial and error, we see that these integers are \(\displaystyle -8\) and 21, so we continue:

\(\displaystyle 2 (6x^{2} -8x+21 x - 28 ) = 0\)

\(\displaystyle 2\left [ (6x^{2} -8x)+( 21 x - 28 ) \right ]= 0\)

\(\displaystyle 2\left [2x (3x -4)+7( 3x - 4) \right ]= 0\)

\(\displaystyle 2(2x+7)(3x-4) = 0\)

One of the factors must be equal to 0, so either:

\(\displaystyle 2x+7 = 0\)

\(\displaystyle 2x = -7\)

\(\displaystyle x = -\frac{7}{2} = -3\frac{1}{2}\)

or

\(\displaystyle 3x-4 = 0\)

\(\displaystyle 3x = 4\)

\(\displaystyle x = \frac{4}{3} = 1 \frac{1}{3 }\)

The correct choice is \(\displaystyle \left \{ -3\frac{1}{2}, 1 \frac{1}{3 } \right \}\).

\(\displaystyle x^{3}+4x^{2}= 16x+ 64\)

\(\displaystyle x^{3}+4x^{2} - 16x - 64 = 16x+ 64 - 16x - 64\)

\(\displaystyle x^{3}+4x^{2} - 16x - 64 = 0\)

\(\displaystyle (x^{3}+4x^{2} )- (16x + 64 )= 0\)

\(\displaystyle x^{2} (x + 4 )- 16 (x + 4 )= 0\)

\(\displaystyle (x^{2} - 16) (x + 4 )= 0\)

\(\displaystyle (x^{2} - 4^{2}) (x + 4 )= 0\)

\(\displaystyle (x - 4 )(x + 4 ) (x + 4 )= 0\)

\(\displaystyle (x - 4 )(x + 4 ) ^{2}= 0\)

By the Zero Product Principle:

\(\displaystyle x- 4 = 0\), in which case \(\displaystyle x = 4\),

or

\(\displaystyle x+ 4 = 0\), in which case \(\displaystyle x = -4\).

The correct choice is \(\displaystyle \left \{ -4, 4\right \}\).