Find the Missing Operator: 9(√6 + √2·√3)² + 4÷2 __ 25(√2 + 2√2)²

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.