If `a=(4lm)/(l+m)`, then the value of `(a+2l)/(a-2l)+(a+2m)/(a-2m)=`

Options

A. `2`
B. `4`
C. `(2lma)/((l+m))`
D. None of these

Answer

To solve the equation \( \frac{a + 2l}{a - 2l} + \frac{a + 2m}{a - 2m} \) given \( a = \frac{4lm}{l + m} \), we will follow these steps:

### Step 1: Substitute the value of \( a \)
We start by substituting \( a \) in the expression:
\[ \frac{\frac{4lm}{l + m} + 2l}{\frac{4lm}{l + m} - 2l} + \frac{\frac{4lm}{l + m} + 2m}{\frac{4lm}{l + m} - 2m} \]

### Step 2: Simplify the first fraction
For the first fraction, we need a common denominator:
\[ \frac{4lm + 2l(l + m)}{4lm - 2l(l + m)} = \frac{4lm + 2l^2 + 2lm}{4lm - 2l^2 - 2lm} \]
This simplifies to:
\[ \frac{6lm + 2l^2}{2lm - 2l^2} = \frac{2(3lm + l^2)}{2(lm - l^2)} = \frac{3lm + l^2}{lm - l^2} \]

### Step 3: Simplify the second fraction
Now we simplify the second fraction in a similar manner:
\[ \frac{4lm + 2m(l + m)}{4lm - 2m(l + m)} = \frac{4lm + 2m^2 + 2lm}{4lm - 2m^2 - 2lm} \]
This simplifies to:
\[ \frac{6lm + 2m^2}{2lm - 2m^2} = \frac{2(3lm + m^2)}{2(lm - m^2)} = \frac{3lm + m^2}{lm - m^2} \]

### Step 4: Combine the two fractions
Now we combine the two simplified fractions:
\[ \frac{3lm + l^2}{lm - l^2} + \frac{3lm + m^2}{lm - m^2} \]
To add these, we need a common denominator:
\[ \frac{(3lm + l^2)(lm - m^2) + (3lm + m^2)(lm - l^2)}{(lm - l^2)(lm - m^2)} \]

### Step 5: Expand the numerator
Expanding the numerator gives:
\[ (3lm^2 - 3l^2m + l^2lm - l^2m^2) + (3lm^2 - 3m^2l + m^2lm - m^2l^2) \]
Combining like terms results in:
\[ 6lm^2 + 6l^2m - 2l^2m^2 \]

### Step 6: Final expression
Thus, the final expression simplifies to:
\[ \frac{6lm^2 + 6l^2m - 2l^2m^2}{(lm - l^2)(lm - m^2)} \]
This can be further simplified if needed.

### Final Answer
The value of \( \frac{a + 2l}{a - 2l} + \frac{a + 2m}{a - 2m} \) is:
\[ \frac{6lm + 6l^2 - 2l^2m}{(lm - l^2)(lm - m^2)} \]