Solve the following equations : `2((x)/(x+1))^(2)-5((x)/(x+1))+2=0\ xne-1`

Answer

To solve the equation \( 2\left(\frac{x}{x+1}\right)^{2} - 5\left(\frac{x}{x+1}\right) + 2 = 0 \) where \( x \neq -1 \), we can follow these steps:

### Step 1: Substitute \( t \)
Let \( t = \frac{x}{x+1} \). Then, we can rewrite the equation in terms of \( t \):
\[ 2t^{2} - 5t + 2 = 0 \]

### Step 2: Factor the quadratic equation
Next, we need to factor the quadratic equation \( 2t^{2} - 5t + 2 = 0 \).
We can rewrite it as:
\[ 2t^{2} - 4t - t + 2 = 0 \]
Grouping the terms:
\[ (2t^{2} - 4t) + (-t + 2) = 0 \]
Factoring by grouping:
\[ 2t(t - 2) - 1(t - 2) = 0 \]
This gives us:
\[ (t - 2)(2t - 1) = 0 \]

### Step 3: Solve for \( t \)
Setting each factor to zero gives us:
1. \( t - 2 = 0 \) → \( t = 2 \)
2. \( 2t - 1 = 0 \) → \( t = \frac{1}{2} \)

### Step 4: Substitute back for \( x \)
Now we substitute back for \( x \) using \( t = \frac{x}{x+1} \).

**For \( t = 2 \):**
\[ \frac{x}{x+1} = 2 \]
Cross-multiplying gives:
\[ x = 2(x + 1) \]
\[ x = 2x + 2 \]
Rearranging gives:
\[ -x = 2 \quad \Rightarrow \quad x = -2 \]

**For \( t = \frac{1}{2} \):**
\[ \frac{x}{x+1} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2x = x + 1 \]
Rearranging gives:
\[ 2x - x = 1 \quad \Rightarrow \quad x = 1 \]

### Step 5: Final solutions
The solutions to the original equation are:
\[ x = -2 \quad \text{and} \quad x = 1 \]