Question

1/√7-2
Rationalise the denominator

Answer 1

Answer lies in the attachment.

Answer 2

Step-by-step explanation:

tex] \bf \underline{Solution-} \\ [/tex]

$\frac{1}{ \sqrt{7} -2}$

$= \frac{1}{ \sqrt{7} - 2} \times \frac{ \sqrt{7} + 2}{ \sqrt{7} + 2 }$

$= \frac{1( \sqrt{7} + 2) }{( \sqrt{7} - 2)( \sqrt{7} + 2) }$

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

$= \frac{ \sqrt{7} + 2 }{7 - 4}$

$= \frac{ \sqrt{7} + 2}{3}$

$\bf{Hence \: the \: denominator \: is \: rationalised.}$