Multiply the question and write your answer in lowest terms,
$\dfrac{{2x - 1}}{{{x^2} + 2x + 4}} \times \dfrac{{{x^4} - 8x}}{{2{x^2} + 5x - 3}} \times \dfrac{{x + 3}}{{{x^2} - 2x}}$

Hint: Here in this question we will try to multiply the given equation step by step then by simplifying the numerator and denominator we will get the desired result , which is to be expressed in the lowest term possible.

Complete step by step answer:
Let us start by multiplying the first two terms, we get
$
  \left[ {\dfrac{{2x - 1}}{{{x^2} + 2x + 4}} \times \dfrac{{{x^4} - 8x}}{{2{x^2} + 5x - 3}}} \right] \times \left\{ {\dfrac{{x + 3}}{{{x^2} - 2x}}} \right\} \\
   = \left[ {\dfrac{{(2x - 1) \times \left( {{x^4} - 8x} \right)}}{{({x^2} + 2x + 4) \times (2{x^2} + 5x - 3)}}} \right] \times \left\{ {\dfrac{{x + 3}}{{{x^2} - 2x}}} \right\} \\
$
In order to make our task easier , we’ll multiply second term to each of the first term elements. We get
\[ = \left[ {\dfrac{{2x \times \left( {{x^4} - 8x} \right) - 1 \times \left( {{x^4} - 8x} \right)}}{{{x^2} \times (2{x^2} + 5x - 3) + 2x \times (2{x^2} + 5x - 3) + 4 \times (2{x^2} + 5x - 3)}}} \right] \times \left\{ {\dfrac{{x + 3}}{{{x^2} - 2x}}} \right\}\]
Now , we’ll multiply and simplify the numerator and denominator both. We get
$
   = \left[ {\dfrac{{2{x^5} - 16{x^2} - {x^4} + 8x}}{{2{x^4} + 5{x^3} - 3{x^2} + 4{x^3} + 10{x^2} - 6x + 8{x^2} + 20x - 12}}} \right] \times \left\{ {\dfrac{{x + 3}}{{{x^2} - 2x}}} \right\} \\
   = \left[ {\dfrac{{2{x^5} - 16{x^2} - {x^4} + 8x}}{{2{x^4} + 9{x^3} + 15{x^2} + 14x - 12}}} \right] \times \left\{ {\dfrac{{x + 3}}{{{x^2} - 2x}}} \right\} \\
$
Now let us multiply the leftover terms using the same procedure as previous i.e. multiplying the second term to each element of the first term and simplifying both numerator and denominator.
On solving , we get
\[ = \left[ {\dfrac{{2{x^5}(x + 3) - 16{x^2}(x + 3) - {x^4}(x + 3) + 8x(x + 3)}}{{2{x^4}({x^2} - 2x) + 9{x^3}({x^2} - 2x) + 15{x^2}({x^2} - 2x) + 14x({x^2} - 2x) - 12({x^2} - 2x)}}} \right]\]
Opening the brackets and simplifying numerator and denominator , we get
$
   = \left[ {\dfrac{{2{x^6} + 6{x^5} - 16{x^3} - 48{x^2} - {x^5} - 3{x^4} + 8{x^2} + 24x}}{{2{x^6} - 4{x^5} + 9{x^5} - 18{x^4} + 15{x^4} - 30{x^3} + 14{x^3} - 28{x^2} - 12{x^2} + 24x}}} \right] \\
   = \left[ {\dfrac{{2{x^6} + 5{x^5} - 16{x^3} - 40{x^2} - 3{x^4} + 24x}}{{2{x^6} + 5{x^5} - 3{x^4} - 16{x^3} - 40{x^2} + 24x}}} \right] \\
$
As both the numerator and denominator are the same they’ll get cancelled out.
= 1

So the lowest possible term in which the given product could be written is 1.

Note: In this question the simplest mistake we can make is multiplication and calculations. So always remember to check your calculations. By these basics you should be able to solve the question. Try to cross verify your multiplication at each stage as they involve a lot of exponents.