Let's solve these problems by completing the square for each given expression. Here’s a step-by-step solution for each part:

1. $x^2 + \frac{7}{3}x + \_\_ = (x + \_\_)^2$

To complete the square:

1. Start with $x^2 + \frac{7}{3}x$.
2. Take half of the coefficient of $x$, which is $\frac{7}{3}$, so $\frac{7}{6}$.
3. Square $\frac{7}{6}$ to get $\frac{49}{36}$.
4. Add $\frac{49}{36}$ to complete the square.

So, we have: $x^2 + \frac{7}{3}x + \frac{49}{36} = \left( x + \frac{7}{6} \right)^2$

2. $x^2 - \frac{5}{7}x + \_\_ = (x + \_\_)^2$

To complete the square:

1. Start with $x^2 - \frac{5}{7}x$.
2. Take half of the coefficient of $x$, which is $-\frac{5}{7}$, so $-\frac{5}{14}$.
3. Square $-\frac{5}{14}$ to get $\frac{25}{196}$.
4. Add $\frac{25}{196}$ to complete the square.

So, we have: $x^2 - \frac{5}{7}x + \frac{25}{196} = \left( x - \frac{5}{14} \right)^2$

3. $x^2 + \frac{2}{3}x + \_\_ = (x + \_\_)^2$

To complete the square:

1. Start with $x^2 + \frac{2}{3}x$.
2. Take half of the coefficient of $x$, which is $\frac{2}{3}$, so $\frac{1}{3}$.
3. Square $\frac{1}{3}$ to get $\frac{1}{9}$.
4. Add $\frac{1}{9}$ to complete the square.

So, we have: $x^2 + \frac{2}{3}x + \frac{1}{9} = \left( x + \frac{1}{3} \right)^2$

Summary of Answers

1. $x^2 + \frac{7}{3}x + \frac{49}{36} = \left( x + \frac{7}{6} \right)^2$
2. $x^2 - \frac{5}{7}x + \frac{25}{196} = \left( x - \frac{5}{14} \right)^2$
3. $x^2 + \frac{2}{3}x + \frac{1}{9} = \left( x + \frac{1}{3} \right)^2$

Would you like any further explanation on completing the square?

Related Questions:

1. How does completing the square relate to solving quadratic equations?
2. Can completing the square be used to derive the quadratic formula?
3. Why is it necessary to add the square of half the coefficient of $x$ in completing the square?
4. How is completing the square useful in graphing parabolas?
5. Are there other methods to solve quadratic equations besides completing the square?

Tip: Completing the square is also helpful for rewriting quadratic equations in vertex form, which makes it easier to identify the vertex of a parabola.