3/2X + 4 x (-1+8X/(5 debajo de esos dos))=1

Let's clarify and solve the equation. The expression seems to be:

\[
\frac{3}{2}X + 4 \times \left(-1 + \frac{8X}{5 - 2}\right) = 1
\]

It looks like the fraction \( \frac{8X}{5 - 2} \) is part of the equation. Let's proceed with this interpretation.

### Step 1: Simplify the expression inside the parentheses

First, simplify the denominator in the fraction:

\[
\frac{8X}{5 - 2} = \frac{8X}{3}
\]

So the equation becomes:

\[
\frac{3}{2}X + 4 \times \left(-1 + \frac{8X}{3}\right) = 1
\]

### Step 2: Distribute the 4 across the terms inside the parentheses

Distribute 4 to both \(-1\) and \(\frac{8X}{3}\):

\[
\frac{3}{2}X + 4(-1) + 4\left(\frac{8X}{3}\right) = 1
\]

This simplifies to:

\[
\frac{3}{2}X - 4 + \frac{32X}{3} = 1
\]

### Step 3: Combine like terms

To combine \(\frac{3}{2}X\) and \(\frac{32X}{3}\), find a common denominator, which is 6.

Convert each term:

\[
\frac{3}{2}X = \frac{9X}{6}
\]
\[
\frac{32X}{3} = \frac{64X}{6}
\]

Now combine them:

\[
\frac{9X}{6} + \frac{64X}{6} = \frac{73X}{6}
\]

So the equation becomes:

\[
\frac{73X}{6} - 4 = 1
\]

### Step 4: Isolate the variable term

Add 4 to both sides:

\[
\frac{73X}{6} = 5
\]

### Step 5: Solve for \(X\)

Multiply both sides by \(\frac{6}{73}\):

\[
X = \frac{30}{73}
\]

Thus, the solution is:

\[
X = \frac{30}{73}
\]

Would you like more details or have any questions?

### Related Questions:
1. How would the solution change if the denominator were different?
2. Can you solve the equation using a graphing method?
3. What is the significance of the common denominator in combining fractions?
4. How would you handle more complex nested fractions in an equation?
5. What is the importance of isolating the variable in an equation?

### Tip:
When dealing with complex fractions, simplifying each part step by step helps in keeping the equation manageable and avoids errors.

Learn how to solve the equation 3/2X + 4 x (-1 + 8X/(5 - 2)) = 1 with a detailed step-by-step solution. This problem involves algebraic equations and fraction operations, guiding through simplification, distribution, and variable isolation. Suitable for advanced high school students.

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