Let $$A = \left\{x \in \mathbb{R}: |x+3| + |x+4| \le 3\right\}$$, $$B = \left\{x \in \mathbb{R}: 3^x \sum_{r=1}^{\infty} \dfrac{3^{x-3}}{10^r} < 3^{-3x}\right\}$$, where $$[t]$$ denotes greatest integer function. Then,

To solve for sets A and B, we will break them down individually and then determine their relationship.

The set is defined as $$A\ =\ \left\{x\in\ R\ :\ \left|x+3\right|+\left|x+4\right|\le3\right\}$$

We evaluate this absolute value inequality by checking three critical regions determined by the roots x = -4 and x = -3.

Case 1: x < -4

Both terms inside the absolute values are negative:

$$-(x + 3) - (x + 4) \le 3 \implies -2x - 7 \le 3 \implies -2x \le 10 \implies x \ge -5$$

In this region, the solution is [-5, -4).

Case 2: -4 <= x <= -3

The first term is negative, the second is positive:

$$-(x + 3) + (x + 4) \le 3 \implies 1 \le 3$$

This is always true. So, the entire interval [-4, -3] is included.

Case 3: x > -3

Both terms are positive:

$$(x + 3) + (x + 4) \le 3 \implies 2x + 7 \le 3 \implies 2x \le -4 \implies x \le -2$$

In this region, the solution is (-3, -2].

Combining the cases:

$$A = [-5, -2]$$

Now Solve for Set(B):

The set is defined as 

First, let's simplify the summation. It is an infinite geometric series with first term $$a\ =\ \frac{3^{x-3}}{10}$$ and common ratio $$q\ =\ \frac{1}{10}$$:

$$\sum_{r=1}^{\infty} \frac{3^{x-3}}{10^r} = 3^{x-3} \left( \frac{1/10}{1 - 1/10} \right) = 3^{x-3} \left( \frac{1/10}{9/10} \right) = \frac{3^{x-3}}{9} = \frac{3^{x-3}}{3^2} = 3^{x-5}$$

Now, substitute this back into the inequality:

$$3^x.3^{x-5}<3^{-3x}$$

Since the base 3 > 1, we can compare the exponents directly:

$$3^{2x-5}<3^{-3x}$$

So, B = (-$$\infty\ $$, 1).

So Finally,

We have:

• $A = [-5, -2]
• $B = (-$$\infty\ $$, 1)

Checking the relationship between the two sets:

Since every value in [-5, -2] is strictly less than 1, all elements of A are contained within B.

Final Result:

$$A \subset B$$

$$B\ =\ \left\{x\ \in\ R\ :\ 3^x\Sigma\ \frac{3^{x-3}}{10^r}<3^{-3x}\right\}$$