If $\left( \frac{x}{x+1} \right)^2 - 5 \left( \frac{x}{x+1} \right) + 6 = 0$, then the positive value of $\left( 1+\frac{1}{x} \right)$ is equal to :

Q: If $\left( \frac{x}{x+1} \right)^2 - 5 \left( \frac{x}{x+1} \right) + 6 = 0$, then the positive value of $\left( 1+\frac{1}{x} \right)$ is equal to :

To solve the given equation and find the positive value of $1 + \frac{1}{x}$, we start by analyzing the equation: \[ \left( \frac{x}{x+1} \right)^2 - 5 \left( \frac{x}{x+1} \right) + 6 = 0 \] Let $ y = \frac{x}{x+1} $. Substituting $ y $ into the equation, we have: \[ y^2 - 5y + 6 = 0 \] This is a quadratic equation in $ y $. We can factor it as: \[ (y - 2)(y - 3) = 0 \] Thus, the solutions for $ y $ are $ y = 2 $ and $ y = 3 $. Since $ y = \frac{x}{x+1} $, we consider each case: 1. **Case 1: $ y = 2 $** \[ \frac{x}{x+1} = 2 \implies x = 2(x + 1) \implies x = 2x + 2 \implies x = -2 \] For $ x = -2 $, we calculate $ 1 + \frac{1}{x} $: \[ 1 + \frac{1}{x} = 1 + \frac{1}{-2} = 1 - \frac{1}{2} = \frac{1}{2} \] 2. **Case 2: $ y = 3 $** \[ \frac{x}{x+1} = 3 \implies x = 3(x + 1) \implies x = 3x + 3 \implies x = -\frac{3}{2} \] For $ x = -\frac{3}{2} $, we calculate $ 1 + \frac{1}{x} $: \[ 1 + \frac{1}{x} = 1 + \frac{1}{-\frac{3}{2}} = 1 - \frac{2}{3} = \frac{1}{3} \] Thus, the positive values of $ 1 + \frac{1}{x} $ are $\frac{1}{2}$ and $\frac{1}{3}$. **Explanation of Options:** - **Option 1 (5):** This value does not match any of the calculated positive values of $ 1 + \frac{1}{x} $. - **Option 2 (6):** This value does not match any of the calculated positive values of $ 1 + \frac{1}{x} $. - **Option 3 ($1\frac{1}{6}$ or $1\frac{1}{5}$):** These values do not match any of the calculated positive values of $ 1 + \frac{1}{x} $. - **Option 4 ($\frac{1}{2}$ or $1\frac{1}{3}$):** This option correctly includes the positive values $\frac{1}{2}$ and $\frac{1}{3}$ that we calculated. Therefore, Option 4 is the correct answer.

Question

Accepted Answer

To solve the given equation and find the positive value of $1 + \frac{1}{x}$, we start by analyzing the equation:

$$
\left( \frac{x}{x+1} \right)^2 - 5 \left( \frac{x}{x+1} \right) + 6 = 0
$$

Let $ y = \frac{x}{x+1} $. Substituting $ y $ into the equation, we have:

$$
y^2 - 5y + 6 = 0
$$

This is a quadratic equation in $ y $. We can factor it as:

$$
(y - 2)(y - 3) = 0
$$

Thus, the solutions for $ y $ are $ y = 2 $ and $ y = 3 $.

Since $ y = \frac{x}{x+1} $, we consider each case:

1. **Case 1: $ y = 2 $**

$$
   \frac{x}{x+1} = 2 \implies x = 2(x + 1) \implies x = 2x + 2 \implies x = -2
   $$

For $ x = -2 $, we calculate $ 1 + \frac{1}{x} $:

$$
   1 + \frac{1}{x} = 1 + \frac{1}{-2} = 1 - \frac{1}{2} = \frac{1}{2}
   $$

2. **Case 2: $ y = 3 $**

$$
   \frac{x}{x+1} = 3 \implies x = 3(x + 1) \implies x = 3x + 3 \implies x = -\frac{3}{2}
   $$

For $ x = -\frac{3}{2} $, we calculate $ 1 + \frac{1}{x} $:

$$
   1 + \frac{1}{x} = 1 + \frac{1}{-\frac{3}{2}} = 1 - \frac{2}{3} = \frac{1}{3}
   $$

Thus, the positive values of $ 1 + \frac{1}{x} $ are $\frac{1}{2}$ and $\frac{1}{3}$.

**Explanation of Options:**

- **Option 1 (5):** This value does not match any of the calculated positive values of $ 1 + \frac{1}{x} $.

- **Option 2 (6):** This value does not match any of the calculated positive values of $ 1 + \frac{1}{x} $.

- **Option 3 ($1\frac{1}{6}$ or $1\frac{1}{5}$):** These values do not match any of the calculated positive values of $ 1 + \frac{1}{x} $.

- **Option 4 ($\frac{1}{2}$ or $1\frac{1}{3}$):** This option correctly includes the positive values $\frac{1}{2}$ and $\frac{1}{3}$ that we calculated.

Therefore, Option 4 is the correct answer.