Question

rationalize the denominator only intelligent people would be able to solve this if you are not intelligent just ignore this question. thank you ​

Answer 1

Required Solution:

$\sf{ \dfrac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} } }$

Here, let :

⪼ $\sf { \sqrt{7} + \sqrt{6} = a }$

⪼ $\sf { \sqrt{13} = b }$

To rationalise the denominator, we'll have to multiply the denominator and the numerator by its rationalising factor. As we supposed (√7 + √6) as a and √13 as b, so the denominator is like ( a - b ). We know that rationalising factor of $\rm { (a -b)}$ is $\rm { (a + b)}$. Therefore, the rationalising factor of $\bigg ( \rm { \dfrac{1}{\sqrt{7} + \sqrt{6} - \sqrt{13} } } \bigg )$ is $( \rm { \sqrt{7} + \sqrt{6} + \sqrt{13} } )$.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{1}{ \sqrt{7} + \sqrt{6} - \sqrt{13} } \times \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ \sqrt{7} + \sqrt{6} + \sqrt{13} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ { \big( \sqrt{7} + \sqrt{6} \big ) }^{2} - {( \sqrt{ 13} )}^{2} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

[ Since, (a - b)(a + b) = a² - b² ]

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ \pink{ {( \sqrt{7}) }^{2} + { (\sqrt{6}) }^{2} + 2( \sqrt{7} \times \sqrt{6} ) } - {( \sqrt{13}) }^{2} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 7 + 6 + 2( \sqrt{42} ) - 13 } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ \cancel{ 13} + 2( \sqrt{42} ) \cancel{- 13} }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 2( \sqrt{42} ) }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now, as we know that rationalising factor of √a is √a since √a × √a is a which is rational. So, rationalising factor of √42 is √42. To rationalise it, we'll multiply √42 with the numerator and the denominator.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{7} + \sqrt{6} + \sqrt{13} }{ 2( \sqrt{42} ) } \times \dfrac{ \sqrt{42} }{ \sqrt{42} } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{42} (\sqrt{7} )+ \sqrt{42} (\sqrt{6} )+ \sqrt{42(} \sqrt{13}) }{ \sqrt{42} (2 \sqrt{42}) } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{294} + \sqrt{252} + \sqrt{546} }{ 2 \times 42 } }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ \sqrt{294} + \sqrt{252} + \sqrt{546} }{ 84} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{ 7\sqrt{6} + 6\sqrt{7} + \sqrt{546} }{ 84} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf { \dfrac{7\sqrt{6}}{84} + \dfrac{6\sqrt{7} }{84} + \dfrac{\sqrt{546} }{ 84} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \boxed{ \sf { \dfrac{\sqrt{6}}{12} + \dfrac{\sqrt{7} }{14} + \dfrac{\sqrt{546} }{ 84} } }$

Points to remember:

• (a + b) and (a - b) are rationalising factors of each other.

• (a + b√x) and (a - b√x) are rationalising factors of each other.

Some Identities:

• (√a)² = a

• √a√b = √ab

• √a/√b = √a/b

• (√a + √b)(√a - √b) = a - b

• (a + √b)(a - √b) = a² - b

• (√a ± √b)² = a ± 2√ab + b

• (√a + √b)(√c + √d) = √ac + √ad + √bc + √bd

Answer 2

Answer:-

That's not true. If you know the correct procedure, you can solve anything related. Only group the denominator into two. That's it.

Refer to the attachment for answer.