Answer:
The expressions in ascending order would be:
[tex]\frac{3x^2+1}{2(x-1)} < \frac{x^2-1}{1-2x} < \frac{x^2}{1-2x} < \frac{2x^2+x}{2}[/tex]
At x = -2
Explanation:
First, we will evaluate the given expressions at x = -2
1- The first expression:
[tex]\frac{x^2}{1-2x}=\frac{(-2)^2}{1-2(-2)}=\frac{4}{5}[/tex]
2- The second expression:
[tex]\frac{x^2-1}{1-2x}=\frac{(-2)^2-1}{1-2(-2)}=\frac{3}{5}[/tex]
3- The third expression:
[tex]\frac{2x^2+x}{2}=\frac{2(-2)^2+(-2)}{2}=3[/tex]
4- The fourth expression:
[tex]\frac{3x^2+1}{2(x-1)}=\frac{3(-2)^2+1}{2(-2-1)}=-\frac{13}{6}[/tex]
Then, we will arrange the values in an ascending order:
[tex]-\frac{13}{6} < \frac{3}{5} < \frac{4}{5} < 3[/tex]
Finally, we arrange the expressions based on the value arrangement:
[tex]\frac{3x^2+1}{2(x-1)} < \frac{x^2-1}{1-2x} < \frac{x^2}{1-2x} < \frac{2x^2+x}{2}[/tex]
Hope this helps :)
