[tex](2x - \frac{1}{3} ) - ( \frac{2}{x} + 1)[/tex]
[tex]2x - \frac{1}{3} - \frac{2}{x} + 1[/tex]
[tex](2x - \frac{2}{x} ) + ( - \frac{1}{3} + 1)[/tex]
[tex] (\frac{2 {x}^{2} }{x} - \frac{2}{x} ) + ( \frac{ - 1}{3} + \frac{3}{3} )[/tex]
[tex] \frac{2 {x}^{2} - 2 }{x} + \frac{ - 2}{3} [/tex]
We want to write the given expression as a single fraction. We will get the fraction:
[tex]\frac{6*x^2 - 4*x - 6}{3*x}[/tex]
So we start with the expression:
[tex](2*x - \frac{1}{3}) - (\frac{2}{x} + 1)[/tex]
First, let's rewrite each of these parenthesis as fractions, we need to have the same denominator in all the terms, then in the first one we can multiply and divide by 3 (and in the second one we can do the same but with x) so we get:
[tex](\frac{3}{3} *2*x - \frac{1}{3}) - (\frac{2}{x} + 1*\frac{x}{x} )\\\\(\frac{6*x}{3} - \frac{1}{3}) - (\frac{2}{x} + \frac{x}{x} )\\\\\frac{6*x - 1}{3} - \frac{2 + x}{x}[/tex]
Now we must multiply and divide the left term by x, and the right term by 3, so we get:
[tex]\frac{x}{x} \frac{6*x - 1}{3} - \frac{2 + x}{x}*\frac{3}{3} \\\\\frac{6*x^2 - x}{3*x} - \frac{6 + 3x}{3*x} = \frac{6*x^2 - x - 3x - 6}{3*x}\\\\\frac{6*x^2 - 4*x - 6}{3*x}[/tex]
This is the fraction in its simplest form.
If you want to learn more about fractions you can read:
https://brainly.com/question/11562149
