Rationalize the denominator of `(1)/(sqrt((a))+sqrt((b))+sqrt((c ))+sqrt((d))+sqrt((b'))+sqrt((c ')))` if `(a)/(d)=(b)/(b')=(c )/(c')`.

Answer

To rationalize the denominator of the expression

\[ \frac{1}{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} + \sqrt{b'} + \sqrt{c'}} \]

given the condition

\[ \frac{a}{d} = \frac{b}{b'} = \frac{c}{c'} \]

let's denote the common ratio as \( k \). Thus, we can express \( a, b, c, d, b', c' \) in terms of \( k \):

1. \( a = kd \)
2. \( b = kb' \)
3. \( c = kc' \)

Now, substituting these values into the denominator:

\[ \sqrt{a} = \sqrt{kd}, \quad \sqrt{b} = \sqrt{kb'}, \quad \sqrt{c} = \sqrt{kc'}, \quad \sqrt{d} = \sqrt{d}, \quad \sqrt{b'} = \sqrt{b'}, \quad \sqrt{c'} = \sqrt{c'} \]

The denominator becomes:

\[ \sqrt{kd} + \sqrt{kb'} + \sqrt{kc'} + \sqrt{d} + \sqrt{b'} + \sqrt{c'} \]

Now, we can factor out \(\sqrt{d}\) from the terms involving \(k\):

\[ = \sqrt{d}(\sqrt{k} + \sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} + 1 + \sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}}) \]

This simplifies to:

\[ = \sqrt{d}(\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} + 1) \]

Now, we can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator. The conjugate is:

\[ \sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} - 1 \]

Thus, we multiply:

\[ \frac{1}{\sqrt{d}(\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} + 1)} \cdot \frac{\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} - 1}{\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} - 1} \]

This gives us:

\[ \frac{\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} - 1}{\sqrt{d}((\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}})^2 - 1^2)} \]

Now, we need to simplify the denominator:

\[ (\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}})^2 - 1 \]

This results in:

\[ k + 4\frac{b'}{d} + 2\sqrt{k \cdot 4\frac{b'}{d}} + \frac{c'}{d} - 1 \]

Finally, we can write the rationalized expression as:

\[ \frac{\sqrt{k} + 2\sqrt{\frac{b'}{d}} + \sqrt{\frac{c'}{d}} - 1}{\sqrt{d}(k + 4\frac{b'}{d} + 2\sqrt{k \cdot 4\frac{b'}{d}} + \frac{c'}{d} - 1)} \]