https://r2.erweima.ai/mathbot/image/17258382260355s0kgqt1.jpg

realiza los siguientes ejercicios correctamente 

Let's solve the two problems presented in the image.

### Problem 7:
We are asked to calculate the value using notable products. The expression is:

\[
16 \times \sqrt{3} \times 5 \times \sqrt{17} \times \sqrt{257} + 1 - \left(\left(\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}}\right)^3 - 9\left(\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}}\right)\right)
\]

**Step 1:** Simplify the product.

Notice that the expression involves a product of square roots and integers. Let's simplify step by step:

\[
\sqrt{3} \times \sqrt{17} \times \sqrt{257} = \sqrt{3 \times 17 \times 257}
\]

Calculate the product inside the square root:

\[
3 \times 17 = 51
\]
\[
51 \times 257 = 13107
\]
\[
\text{Thus, } \sqrt{3} \times \sqrt{17} \times \sqrt{257} = \sqrt{13107}
\]

The expression inside becomes:

\[
16 \times \sqrt{13107} + 1 - \left(\left(\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}}\right)^3 - 9\left(\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}}\right)\right)
\]

**Step 2:** Simplify the expression involving square roots and powers.

Notice the sum inside the cube:

\[
\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}}
\]

This sum is a well-known identity equal to \(\sqrt{2}\), because:

\[
\sqrt{a + \sqrt{b}} + \sqrt{a - \sqrt{b}} = \sqrt{2a + 2\sqrt{a^2 - b}}
\]

Thus,

\[
\sqrt{3 + 2\sqrt{2}} + \sqrt{3 - 2\sqrt{2}} = \sqrt{4} = 2
\]

Now substitute this into the original expression:

\[
\text{Expression } = 16 \times \sqrt{13107} + 1 - (2^3 - 9 \times 2)
\]

\[
= 16 \times \sqrt{13107} + 1 - (8 - 18)
\]

\[
= 16 \times \sqrt{13107} + 1 - (-10)
\]

\[
= 16 \times \sqrt{13107} + 11
\]

So, the value is:

\[
16 \times \sqrt{13107} + 11
\]

### Problem 8:
We need to find \( K \) such that:

\[
(a + 3b)^2 + (a + b)^2 + Kb^2
\]

is a Perfect Square Trinomial (TCP in Spanish).

**Step 1:** Expand the squares.

\[
(a + 3b)^2 = a^2 + 6ab + 9b^2
\]

\[
(a + b)^2 = a^2 + 2ab + b^2
\]

Add these expressions:

\[
a^2 + 6ab + 9b^2 + a^2 + 2ab + b^2 + Kb^2
\]

\[
= 2a^2 + 8ab + (10 + K)b^2
\]

**Step 2:** For this expression to be a perfect square trinomial, we need:

\[
2a^2 + 8ab + (10 + K)b^2 = (ca + db)^2
\]

This gives us:

\[
c^2 = 2, \quad 2cd = 8, \quad d^2 = 10 + K
\]

From \( c^2 = 2 \), \( c = \sqrt{2} \).

From \( 2cd = 8 \), substitute \( c = \sqrt{2} \):

\[
2\sqrt{2}d = 8 \quad \Rightarrow \quad d = \frac{8}{2\sqrt{2}} = 2\sqrt{2}
\]

Then, \( d^2 = 8 \), so:

\[
8 = 10 + K \quad \Rightarrow \quad K = -2
\]

Thus, the value of \( K \) is \(-2\).

### Summary:
- The value from Problem 7 is \( 16 \times \sqrt{13107} + 11 \).
- The value of \( K \) from Problem 8 is \( -2 \).

Would you like more details on any step, or have any questions?

---

**Relative Questions:**
1. What is the significance of notable products in simplifying expressions?
2. How would the value change in Problem 7 if the coefficients were different?
3. Can you derive an alternative method for simplifying square root expressions?
4. What conditions must be met for a quadratic expression to be a perfect square trinomial?
5. How does changing the coefficients of \( b^2 \) in Problem 8 affect the value of \( K \)?

**Tip:** When solving expressions involving square roots and powers, always look for patterns or identities that can simplify the calculations.

Aplicando productos notables, calcular el valor de 16√3 × 5 × 17 × √257 + 1 y hallar el valor de K si (a + 3b)^2 + (a + b)^2 + Kb^2 es un TCP.

Explore how to calculate expressions involving notable products and find the value of K in a perfect square trinomial. This problem covers advanced algebraic concepts such as square roots, algebraic identities, and perfect square trinomials. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation