Solve 
1
x
+
2
x
+
1
=
3
x
+
2
x
1
​	
 + 
x+1
2
​	
 = 
x+2
3
​	
 

Let's solve the equation step by step:

\[
\frac{1}{x} + \frac{2}{x+1} = \frac{3}{x+2}
\]

### Step 1: Find the least common denominator (LCD)
The LCD of the fractions \(\frac{1}{x}\), \(\frac{2}{x+1}\), and \(\frac{3}{x+2}\) is \(x(x+1)(x+2)\).

### Step 2: Multiply each term by the LCD to eliminate the fractions
Multiply both sides of the equation by the LCD:

\[
x(x+1)(x+2) \cdot \frac{1}{x} + x(x+1)(x+2) \cdot \frac{2}{x+1} = x(x+1)(x+2) \cdot \frac{3}{x+2}
\]

This simplifies to:

\[
(x+1)(x+2) + 2x(x+2) = 3x(x+1)
\]

### Step 3: Expand and simplify the equation
Now expand each term:

\[
(x^2 + 3x + 2) + 2x^2 + 4x = 3x^2 + 3x
\]

Combine like terms:

\[
x^2 + 3x + 2 + 2x^2 + 4x = 3x^2 + 3x
\]

\[
3x^2 + 7x + 2 = 3x^2 + 3x
\]

### Step 4: Move all terms to one side of the equation
Subtract \(3x^2 + 3x\) from both sides:

\[
3x^2 + 7x + 2 - 3x^2 - 3x = 0
\]

This simplifies to:

\[
4x + 2 = 0
\]

### Step 5: Solve for \(x\)
Subtract 2 from both sides:

\[
4x = -2
\]

Divide by 4:

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

So, the solution is \(x = -\frac{1}{2}\).

Would you like more details or have any questions? Here are some related problems you might find useful:

1. Solve \(\frac{1}{x} + \frac{3}{x+2} = \frac{5}{x+4}\).
2. Solve \(\frac{2}{x} + \frac{4}{x+3} = \frac{6}{x+5}\).
3. Simplify and solve \(\frac{1}{x-1} + \frac{2}{x+2} = \frac{3}{x+3}\).
4. Solve \(\frac{2}{x+1} + \frac{1}{x-2} = \frac{3}{x+3}\).
5. Solve the quadratic equation \(x^2 - 4x + 4 = 0\).
6. Find the roots of \(x^2 + 6x + 9 = 0\).
7. Solve \(\frac{1}{x-2} + \frac{3}{x+2} = \frac{4}{x+4}\).
8. Solve the system of equations \(x + y = 5\) and \(\frac{1}{x} + \frac{1}{y} = \frac{1}{2}\).

**Tip:** For rational equations, always check that your solution does not make any denominator zero, as that would make the solution invalid.

Learn how to solve the rational equation (1/x) + (2/(x+1)) = (3/(x+2)) step-by-step with a detailed solution. This problem covers key concepts in solving rational equations, including finding the least common denominator (LCD). Suitable for high school students (Grades 10-12).

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