Solve the equations:
Q. `sqrt((2x^(2)+1)/(x^(2)-1))+6sqrt((x^(2)-1)/(2x^(2)+1))=5`.

Answer

To solve the equation

\[ \sqrt{\frac{2x^2 + 1}{x^2 - 1}} + 6\sqrt{\frac{x^2 - 1}{2x^2 + 1}} = 5, \]

we will follow these steps:

### Step 1: Substitute for Simplification
Let

\[ y = \sqrt{\frac{2x^2 + 1}{x^2 - 1}}. \]

Then,

\[ \sqrt{\frac{x^2 - 1}{2x^2 + 1}} = \frac{1}{y}. \]

Substituting these into the original equation gives:

\[ y + 6 \cdot \frac{1}{y} = 5. \]

### Step 2: Clear the Fraction
Multiply through by \(y\) to eliminate the fraction:

\[ y^2 + 6 = 5y. \]

### Step 3: Rearrange into Standard Quadratic Form
Rearranging the equation yields:

\[ y^2 - 5y + 6 = 0. \]

### Step 4: Factor the Quadratic
Now, we factor the quadratic equation:

\[ (y - 2)(y - 3) = 0. \]

### Step 5: Solve for \(y\)
Setting each factor to zero gives:

\[ y - 2 = 0 \quad \Rightarrow \quad y = 2, \]
\[ y - 3 = 0 \quad \Rightarrow \quad y = 3. \]

### Step 6: Substitute Back for \(x\)
Recall that

\[ y = \sqrt{\frac{2x^2 + 1}{x^2 - 1}}. \]

#### Case 1: \(y = 2\)

\[ 2 = \sqrt{\frac{2x^2 + 1}{x^2 - 1}}. \]

Squaring both sides:

\[ 4 = \frac{2x^2 + 1}{x^2 - 1}. \]

Cross-multiplying gives:

\[ 4(x^2 - 1) = 2x^2 + 1. \]

Expanding and rearranging:

\[ 4x^2 - 4 = 2x^2 + 1 \quad \Rightarrow \quad 2x^2 - 5 = 0. \]

Solving for \(x^2\):

\[ x^2 = \frac{5}{2} \quad \Rightarrow \quad x = \pm \sqrt{\frac{5}{2}}. \]

#### Case 2: \(y = 3\)

\[ 3 = \sqrt{\frac{2x^2 + 1}{x^2 - 1}}. \]

Squaring both sides:

\[ 9 = \frac{2x^2 + 1}{x^2 - 1}. \]

Cross-multiplying gives:

\[ 9(x^2 - 1) = 2x^2 + 1. \]

Expanding and rearranging:

\[ 9x^2 - 9 = 2x^2 + 1 \quad \Rightarrow \quad 7x^2 - 10 = 0. \]

Solving for \(x^2\):

\[ x^2 = \frac{10}{7} \quad \Rightarrow \quad x = \pm \sqrt{\frac{10}{7}}. \]

### Final Solutions
Thus, the solutions for \(x\) are:

\[ x = \sqrt{\frac{5}{2}}, \quad x = -\sqrt{\frac{5}{2}}, \quad x = \sqrt{\frac{10}{7}}, \quad x = -\sqrt{\frac{10}{7}}. \]