To solve the given problem and **determine whether it is an equation or inequality,** we need to simplify each of the algebraic expressions.

We can deal with each of them separately and simplify them. **However, in this case, a more efficient way is be to deal with the more complex parts of these expressions separately**, that is - the expressions in parentheses - the expressions with roots.

__It's important to emphasize that in general we want to solve without a calculator,__ using only our algebraic tools and laws of exponents. Let's begin:

__a.__ **We'll start** with the problematic part **in the left expression**:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3}$(We'll focus on the expression inside the parentheses first and then move outwards),

**Let's recall** two laws of exponents:

__a.1:__ Defining a root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

__a.2:__ The law of applying exponents to parentheses containing a product, **but in the opposite direction:**

$x^n\cdot y^n= (x\cdot y)^n$

**Usually** we replace roots with exponents, but for now **we won't do that**. We'll just understand that according to the law of defining roots as an exponent mentioned in **a.1**, __the root is actually an exponent and therefore all laws of exponents apply to it, __especially the law of exponents mentioned in **a.2** , So we'll apply this understanding to the expression in question:

$\sqrt{6}+\sqrt{2}\cdot\sqrt{3} = \sqrt{6}+\sqrt{2\cdot3}= \sqrt{6}+\sqrt{6}$

In the first stage, we notice that the second term (i.e., the product of roots) is actually a product between two terms raised to **the same exponent (which is half the power of the square root)**. Therefore, **according to the law of exponents mentioned in a.2 **, we can combine the bases of the terms as a product **with the same exponent **, and in the next stage we simplify the expression under the root,

From here we notice that** we can simplify this expression** by a using common factor:

$\sqrt{6}+\sqrt{6}=(1+1)\sqrt{6}=2\cdot\sqrt{6}$

**We used the commutative property of multiplication** to move the common factor we took out ($\sqrt{6}$ ) we put it on **the right of the parentheses (instead of to their left)** so our expression with be clearer.

Now,** we'll return **to the **original** expression in the problem (i.e., the expression on the left) and calculate in full, using the simplification from above:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1}$

We **substitute** what we calculated above in place of the expression in parentheses and write the division operation for the last term on the left as a fraction.

**In the next stage** let's recall another law of exponents:

a.3: The law of dividing exponents with equal bases:

$\frac{a^m}{a^n}=a^{m-n}$And let's recall again the law of exponents mentioned above in a.2, that is, the law of an exponent applied to parentheses containing a product, **but in the normal direction.**

Let's apply these two laws of exponents to the expression we got in the last stage:

$9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =9\cdot2^2\cdot(\sqrt{6})^2+2^{2-1}=9\cdot4\cdot6+2^1$

In the first stage we apply the law of exponents mentioned in **a.2** above and apply the exponent to **each of the multiplied terms in the parentheses. **

Then, we apply the law of exponents mentioned in **a.3** to the second term on the left. To make things clearer, we put the root in parentheses, **but this just for convenience**.

In the next stage we square the square root __while remembering that these are actually two inverse operations and therefore cancel each other out__ and we simplify the rest of the terms.

**Let's finish **the calculation. We got that the expression is:

$9\cdot4\cdot6+2^1 =216+2=218$

**Let's summarize**:

We got that the left expression is:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2= 9\cdot(2\cdot\sqrt{6})^2+\frac{2^2}{2^1} =\\
9\cdot4\cdot6+2^1 =218$

__b.__ Let's continue to the expression on the right and as before,we'll start with the problematic part, that is, the expression inside the parentheses with the roots:

$\sqrt{2}+2\sqrt{2}$**Just as in the previous part**, we can factor this expression:

$\sqrt{2}+2\sqrt{2} =(1+2)\sqrt{2}=3\cdot\sqrt{2}$

Again we use the commutative property of multiplication and the common factor-$\sqrt{2}$ we choose **to take out outside the parentheses - to their right.**

Next, we simplify the expression in parentheses.

**We'll return to the full expression on the right** and substitutewhat we got :

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2$

We'll continue and simplify this expression. Again, we wil apply the law of exponents mentioned earlier in **a.2 (in its normal direction)** and we __keep in mind that the square root and the square exponent are inverse operations and therefore cancel each other out:__

$5^2\cdot(3\cdot\sqrt{2})^2 =5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2$

First, according to the law of exponents mentioned in **a.2 **we apply the exponent to each of the multiplied terms in parentheses, and then we simplify the resulting expression while we keep in mind __ that the square root and square exponent are inverse operations.__

**Let's finish **the calculation:

$25\cdot9\cdot2=450$**And to summarize** , we got that:

$5^2\cdot(\sqrt{2}+2\sqrt{2})^2 =5^2\cdot(3\cdot\sqrt{2})^2 =\\
5^2\cdot3^2 \cdot(\sqrt{2})^2=25\cdot9\cdot2 =450$

**Now let's return to the original problem** and substitute what we got in a and b:

$9\cdot(\sqrt{6}+\sqrt{2}\cdot\sqrt{3})^2+2^2:2\text{ }\text{\textcolor{red}{\_\_}}\text{ }5^2\cdot(\sqrt{2}+2\sqrt{2})^2 \\
\downarrow\\
218\text{ }\text{\textcolor{red}{\_\_}}\text{ }450$**Therefore it's clear that this is not an equality but an inequality **and that the expression on the left is smaller than the expression on the right ,**that is:**

218\text{ }\text{\textcolor{red}{<}}\text{ }450 __Therefore the correct answer is answer a.__