I see you've uploaded an image containing six mathematical inequalities. Let's go over them step by step to solve each one. The equations are as follows:

1. $\frac{x + 2}{4 - 2x} \geq 1 - x$
2. $\frac{x - 2}{x^2} \leq \frac{x + 1}{x}$
3. $|2 - x| + |3 - 2x| \leq 3$
4. $|x + 1|^2 + 2|x + 2| \geq 2$
5. $2x + 3 \geq |4x + 5|$
6. $|x| + 3x \leq 2$

Let's start by solving the first inequality:

1. $\frac{x + 2}{4 - 2x} \geq 1 - x$

We'll proceed with the following steps:

• Combine terms on one side.
• Eliminate the denominator by multiplying through.
• Solve for $x$.

I'll work through this now.The solution to the inequality $\frac{x + 2}{4 - 2x} \geq 1 - x$ is:

$\left(\frac{7 - \sqrt{33}}{4} \leq x < 2 \right) \cup \left( \frac{7 + \sqrt{33}}{4} \leq x \right)$

This means:

• $x$ lies between $\frac{7 - \sqrt{33}}{4}$ and 2, or
• $x \geq \frac{7 + \sqrt{33}}{4}$.

Would you like me to move on to solving the second inequality?

Here are five related questions you might consider:

1. What happens if you change the inequality to a strict inequality (i.e., $>$)?
2. How would this inequality change if you multiplied through by $4 - 2x$?
3. What are the critical points where the expression changes sign?
4. Can we solve this inequality graphically?
5. How would you check the solution using substitution?

Tip: Always check for restrictions on variables (like denominators that cannot be zero) when solving inequalities.