https://r2.erweima.ai/mathbot/image/1726072714130idptfqjo.jpg

Let's solve the exercises in the image. The task is to write the calculations in their irreducible form, with detailed steps and no calculator. I'll solve each step-by-step:

---

**Exercise A:**

\[
A = \frac{5}{12} - \frac{3}{8}
\]

To subtract these fractions, we need a common denominator. The least common denominator (LCD) of 12 and 8 is 24. Let's rewrite both fractions:

\[
\frac{5}{12} = \frac{10}{24}, \quad \frac{3}{8} = \frac{9}{24}
\]

Now, subtract the numerators:

\[
\frac{10}{24} - \frac{9}{24} = \frac{1}{24}
\]

So, the irreducible form of \( A \) is:

\[
A = \frac{1}{24}
\]

---

**Exercise B:**

\[
B = \frac{-17}{-15} - \frac{-5}{9}
\]

Simplify the first fraction:

\[
\frac{-17}{-15} = \frac{17}{15}
\]

Now, we need a common denominator. The LCD of 15 and 9 is 45. Let's rewrite both fractions:

\[
\frac{17}{15} = \frac{51}{45}, \quad \frac{-5}{9} = \frac{-25}{45}
\]

Now, subtract:

\[
\frac{51}{45} - \frac{-25}{45} = \frac{51 + 25}{45} = \frac{76}{45}
\]

This fraction is already irreducible, so:

\[
B = \frac{76}{45}
\]

---

**Exercise C:**

\[
C = \frac{1}{5} - \frac{11}{25}
\]

We need a common denominator. The LCD of 5 and 25 is 25. Let's rewrite the fractions:

\[
\frac{1}{5} = \frac{5}{25}
\]

Now, subtract:

\[
\frac{5}{25} - \frac{11}{25} = \frac{5 - 11}{25} = \frac{-6}{25}
\]

So, the irreducible form of \( C \) is:

\[
C = \frac{-6}{25}
\]

---

**Exercise D:**

\[
D = \frac{6}{-7} \times \frac{14}{24}
\]

First, simplify the fractions:

\[
\frac{14}{24} = \frac{7}{12}, \quad \frac{6}{-7} = \frac{-6}{7}
\]

Now, multiply:

\[
\frac{-6}{7} \times \frac{7}{12} = \frac{-6 \times 7}{7 \times 12} = \frac{-42}{84}
\]

Simplify \( \frac{-42}{84} \):

\[
\frac{-42}{84} = \frac{-1}{2}
\]

So, the irreducible form of \( D \) is:

\[
D = \frac{-1}{2}
\]

---

**Exercise E:**

\[
E = \frac{5}{4} \div \frac{5}{5}
\]

Dividing by a fraction is the same as multiplying by its reciprocal. So:

\[
E = \frac{5}{4} \times \frac{5}{5} = \frac{5 \times 5}{4 \times 5} = \frac{25}{20}
\]

Now, simplify \( \frac{25}{20} \):

\[
\frac{25}{20} = \frac{5}{4}
\]

So, the irreducible form of \( E \) is:

\[
E = \frac{5}{4}
\]

---

Do you want more details on any of these steps, or do you have further questions?

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Here are 5 related questions:

1. How do you find the least common denominator (LCD) of two fractions?
2. How do you convert a fraction to its simplest form?
3. What is the general rule for subtracting fractions?
4. How can you multiply fractions and simplify the result?
5. What is the rule for dividing two fractions?

**Tip:** Always simplify fractions after each operation to ensure the result is in its simplest form.

Write the following calculations in their irreducible form by detailing the steps without using a calculator.

Learn how to simplify fractions and perform operations like subtraction, multiplication, and division with fractions. This exercise guides you through writing calculations in irreducible form with detailed steps, suitable for students in Grades 6-8.

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