https://file.aiquickdraw.com/mathbot/file/1729374086590b46kdk11.jpg

I'll go through the exercises one by one and provide detailed steps for solving them. Let’s begin with **Exercice 1**:

### **Exercice 1:**

We are asked to calculate the values of the following expressions, detailing the intermediate steps:

**A = \( 5\sqrt{81} - 3\sqrt{25} + 7(\sqrt{-2})^2 \)**

1. \( \sqrt{81} = 9 \) and \( \sqrt{25} = 5 \).
2. For \( (\sqrt{-2})^2 \), \( \sqrt{-2} \) is not defined in the real number system, but assuming we work with complex numbers:
   - \( \sqrt{-2} = i\sqrt{2} \), and \( (\sqrt{-2})^2 = -2 \).
   
So:
\[
A = 5(9) - 3(5) + 7(-2) = 45 - 15 - 14 = 16.
\]

**B = \( \frac{\sqrt{32} \times \sqrt{2}}{\sqrt[3]{8}} - 8 \)**

1. \( \sqrt{32} \times \sqrt{2} = \sqrt{32 \times 2} = \sqrt{64} = 8 \).
2. \( \sqrt[3]{8} = 2 \), so:
\[
B = \frac{8}{2} - 8 = 4 - 8 = -4.
\]

**C = \( \sqrt{7} \times \frac{\sqrt{15}}{\sqrt{35}} \times \sqrt{3} \)**

1. \( \frac{\sqrt{15}}{\sqrt{35}} = \sqrt{\frac{15}{35}} = \sqrt{\frac{3}{7}} \).
2. So, \( C = \sqrt{7} \times \sqrt{\frac{3}{7}} \times \sqrt{3} = \sqrt{3 \times 3} = \sqrt{9} = 3 \).

**D = \( \left( \frac{\sqrt{11}}{\sqrt{7}} \right)^{-1} \times \left( \sqrt{11} \right)^2 \)**

1. \( \left( \frac{\sqrt{11}}{\sqrt{7}} \right)^{-1} = \frac{\sqrt{7}}{\sqrt{11}} \).
2. \( \left( \sqrt{11} \right)^2 = 11 \).
3. So, \( D = \frac{\sqrt{7}}{\sqrt{11}} \times 11 = 11 \times \frac{\sqrt{7}}{\sqrt{11}} = \frac{11\sqrt{7}}{\sqrt{11}} = \sqrt{77} \).

**E = \( \frac{\sqrt{32} + \sqrt{10} \times 5 \times 3 - \sqrt{100}}{2} \)**

1. \( \sqrt{32} = 4\sqrt{2} \), \( \sqrt{100} = 10 \), and \( \sqrt{10} \times 5 \times 3 = 15\sqrt{10} \).
2. So, \( E = \frac{4\sqrt{2} + 15\sqrt{10} - 10}{2} \).

This expression cannot be simplified much further. The final simplified form is:
\[
E = \frac{4\sqrt{2} + 15\sqrt{10} - 10}{2}.
\]

**F = \( \left( \frac{2}{3} \right)^{-3} \times \left( \frac{3}{2} \right)^{-3} \)**

1. \( \left( \frac{2}{3} \right)^{-3} = \left( \frac{3}{2} \right)^3 = \frac{27}{8} \).
2. \( \left( \frac{3}{2} \right)^{-3} = \left( \frac{2}{3} \right)^3 = \frac{8}{27} \).
3. So:
\[
F = \frac{27}{8} \times \frac{8}{27} = 1.
\]

**G = \( (\sqrt{3} + \sqrt{2}) \times (\sqrt{3} - \sqrt{2}) \)**

1. This is a difference of squares: 
\[
G = \sqrt{3}^2 - \sqrt{2}^2 = 3 - 2 = 1.
\]

**H = \( 1 - \left[ 1 - \left( \frac{15}{22} + \frac{25}{35} + \frac{3}{11} \right) \right] \times \frac{3}{55} \)**

First, simplify the fractions inside:
- \( \frac{15}{22} = \frac{15}{22} \),
- \( \frac{25}{35} = \frac{5}{7} \), and
- \( \frac{3}{11} = \frac{3}{11} \).

Adding these gives:
\[
\frac{15}{22} + \frac{5}{7} + \frac{3}{11}.
\]

To solve this, we’ll need a common denominator (which is 154), but let's move forward to **Exercice 2** for now. 

---

Would you like the details for **Exercice 2** next, or any clarifications on **Exercice 1**?

Learn how to solve math problems involving radical simplification, algebraic expansion, rationalization of denominators, and scientific notation. Detailed step-by-step solutions for each exercise suitable for students in Grades 10-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation