If `4/(2+sqrt3+sqrt7)=sqrta+sqrtb-sqrtc`,then which of the following can be true -

Options

A. a=1, b=4/3, c=7/3
B. `a=1, b=2/3, c=7/9
C. a=2/3, b=1, c=7/3
D. a= 7/9, b=4/3, c=1

Answer

To solve the equation \( \frac{4}{2 + \sqrt{3} + \sqrt{7}} = \sqrt{a} + \sqrt{b} - \sqrt{c} \), we will follow these steps:

### Step 1: Rationalize the Denominator
We start with the expression on the left side:
\[ \frac{4}{2 + \sqrt{3} + \sqrt{7}} \]
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \( 2 + \sqrt{3} - \sqrt{7} \):
\[ \frac{4(2 + \sqrt{3} - \sqrt{7})}{(2 + \sqrt{3} + \sqrt{7})(2 + \sqrt{3} - \sqrt{7})} \]

### Step 2: Simplify the Denominator
Using the difference of squares formula \( (a + b)(a - b) = a^2 - b^2 \):
\[ (2 + \sqrt{3})^2 - (\sqrt{7})^2 \]
Calculating \( (2 + \sqrt{3})^2 \):
\[ = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]
Now calculating \( (\sqrt{7})^2 \):
\[ = 7 \]
Thus, the denominator becomes:
\[ (7 + 4\sqrt{3}) - 7 = 4\sqrt{3} \]

### Step 3: Substitute Back
Now substituting back into our expression:
\[ \frac{4(2 + \sqrt{3} - \sqrt{7})}{4\sqrt{3}} = \frac{2 + \sqrt{3} - \sqrt{7}}{\sqrt{3}} \]

### Step 4: Separate the Terms
We can separate the terms in the numerator:
\[ \frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{\sqrt{3}} - \frac{\sqrt{7}}{\sqrt{3}} = \frac{2}{\sqrt{3}} + 1 - \frac{\sqrt{7}}{\sqrt{3}} \]

### Step 5: Combine into a Single Expression
Now we can express this as:
\[ 1 + \frac{2 - \sqrt{7}}{\sqrt{3}} \]

### Step 6: Express in the Form \( \sqrt{a} + \sqrt{b} - \sqrt{c} \)
To express \( \frac{2 - \sqrt{7}}{\sqrt{3}} \) in the form \( \sqrt{b} - \sqrt{c} \), we can multiply the numerator and denominator by \( \sqrt{3} \):
\[ = \sqrt{3} \cdot \frac{2 - \sqrt{7}}{3} = \sqrt{3} \cdot \left( \frac{2}{3} - \frac{\sqrt{7}}{3} \right) \]
Thus, we can identify:
- \( a = 1 \)
- \( b = \frac{4}{3} \)
- \( c = \frac{7}{3} \)

### Conclusion
Thus, we find that:
- \( a = 1 \)
- \( b = \frac{4}{3} \)
- \( c = \frac{7}{3} \)

The correct option is \( a \).