Question

rationalize the denominator 1/underroot6-underroot7​

Answer 1

$\large\underline{\sf{Solution-}}$

Given

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{6} - \sqrt{7} }$

On rationalizing the denominator means, we multiply and divide the given expression by conjugate of denominator, means with opposite sign in binomial.

So,

$\rm \: = \: \:\dfrac{1}{ \sqrt{6} - \sqrt{7} } \times \dfrac{ \sqrt{6} + \sqrt{7} }{ \sqrt{6} + \sqrt{7} }$

We know,

$\red{ \boxed{ \rm{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

So, we get

$\rm \: = \: \:\dfrac{ \sqrt{6} + \sqrt{7} }{ {( \sqrt{6} )}^{2} - {( \sqrt{7} )}^{2} }$

$\rm \: = \: \:\dfrac{ \sqrt{6} + \sqrt{7} }{6 - 7}$

$\rm \: = \: \:\dfrac{ \sqrt{6} + \sqrt{7} }{- 1}$

$\rm \: = \: \: - \sqrt{6} - \sqrt{7}$

Hence,

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{6} - \sqrt{7} } = - \sqrt{6} - \sqrt{7}$

Additional Information :-

Let's solve more problems of same type!!

Question :- Rationalize the denominator of following :

$\red{\rm :\longmapsto\:1) \: \: \: \dfrac{ \sqrt{5} }{ \sqrt{3} + \sqrt{2} }}$

Solution :

${\rm :\longmapsto\:\: \: \: \dfrac{ \sqrt{5} }{ \sqrt{3} + \sqrt{2} }}$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{ \sqrt{5} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

We know that

$\red{ \boxed{ \rm{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

So, we get

$\rm \: = \: \:\dfrac{ \sqrt{5}( \sqrt{3} - \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} }$

$\rm \: = \: \:\dfrac{ \sqrt{15} - \sqrt{10} }{3 - 2}$

$\rm \: = \: \:\dfrac{ \sqrt{15} - \sqrt{10} }{1}$

$\rm \: = \: \: \sqrt{15} - \sqrt{10}$

Hence,

$\red{\rm :\longmapsto\: \: \: \dfrac{ \sqrt{5} }{ \sqrt{3} + \sqrt{2}} = \sqrt{15} - \sqrt{10} }$

$\green{\rm :\longmapsto\:2) \: \: \: \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }}$

So, on rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

We know,

$\red{ \boxed{ \rm{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$\rm \: = \: \:\dfrac{ {( \sqrt{3} - \sqrt{2} )}^{2} }{ {( \sqrt{3})}^{2} - {( \sqrt{2} )}^{2} }$

We know,

$\red{ \boxed{ \rm{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

$\rm \: = \: \:\dfrac{3 + 2 - 2 \sqrt{6} }{3 - 2}$

$\rm \: = \: \:\dfrac{5 - 2 \sqrt{6} }{1}$

$\rm \: = \: \:5 - 2 \sqrt{6}$

Hence,

$\green{\rm :\longmapsto\: \: \: \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } = 5 - 2 \sqrt{6} }$

Answer 2

$\begin{gathered}\large {\boxed{\sf{\mid{\overline {\underline {\star QUESTION ::}}}\mid}}}\end{gathered}$

rationalize the denominator 1/underroot6-underroot7

$\Huge{\textbf{\textsf{\color{navy}{An}{\purple{sW}{\pink{eR{\color{pink}{:-}}}}}}}}$

$</p><p> \frac{1}{ \sqrt{6} - \sqrt{7} } \\ \\ = \frac{1}{ \sqrt{6} - \sqrt{7} } \times \frac{ \sqrt{6} + \sqrt{7} }{ \sqrt{6} + \sqrt{7} } \\ \\ = \frac{ \sqrt{6} + \sqrt{7} }{6 - 7} \\ \\ = \frac{ \sqrt{6} + \sqrt{7} }{ - 1}$

$- ( \sqrt{6} + \sqrt{7} ) \\ \\ = - \sqrt{6} - \sqrt{7}$

hope its helpful⭐✨⭐✨

⟼$\huge\sf\underline\red{answered\:by\:KIMU}$