Question

Rationalize the denominator of 1/√7+√6−√13

Answer 1

1/(√7+√6-√13)

Rationalising factor = √7+√6+√13

→ 1/(√7+√6-√13) × √7+√6+√13/(√7+√6+√13)

→ √7+√6+√13/(√7+√6-√13)(√7+√6+√13)

→ √7+√6+√13/(√7+√6)²-(√13)²

→ √7+√6+√13/(√7²+√6²+2(√7)(√6))-13

→ √7+√6+√13/13+2√42-13

→ √7+√6+√13/2√42

The denominator is still irrational. So we have to rationalise it further.

Now rationalising factor = √42

→ √7+√6+√13/2√42 × √42/√42

→ √42(√7+√6+√13)/2(√42)²

→ √42×7+√42×6+√42×13/2(42)

→ (7√6+6√7+√546)/84

→ 7√6/84 + 6√7/84 + √546/84

→ √6/12 + √7/14 + √546/84

Now the denominator is rationalised.

Answer 2

Answer:

$\frac{ \sqrt{6} }{12} + \frac{ \sqrt{7} }{14} + \frac{ \sqrt{546} }{84}$

Step-by-step explanation:

$\frac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} }$

$\frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{( \sqrt{7} + { \sqrt{6} )}^{2} - 13}$

$\frac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{2 \sqrt{42} }$

$\frac{ \sqrt{6} }{12} + \frac{ \sqrt{7} }{14} + \frac{ \sqrt{546} }{84}$