Mr Daniels Maths
Rationalising the Denominator

Set 1

Set 2

Set 3

Q1) \( 1 \over{ \sqrt{ 5}} \) = [ \(\sqrt{5}\over{ 5} \)]

Q1) \( 5 \over{ \sqrt{ 8}} \) = [ \(5\sqrt{2}\over{4} \)]

Q1) \({1} \over{ 5 + \sqrt 5} \) = [ \({ 5 - \sqrt 5} \over20 \)]

Q2) \( 1 \over{ \sqrt{ 2}} \) = [ \(\sqrt{2}\over{ 2} \)]

Q2) \( 2 \over{ \sqrt{ 6}} \) = [ \(\sqrt{6}\over{3} \)]

Q2) \({1} \over{ 4 - \sqrt 2} \) = [ \({ 4 + \sqrt 2} \over14 \)]

Q3) \( 1 \over{ \sqrt{ 6}} \) = [ \(\sqrt{6}\over{ 6} \)]

Q3) \( 5 \over{ \sqrt{ 10}} \) = [ \(\sqrt{10}\over{2} \)]

Q3) \({1} \over{ 2 - \sqrt 2} \) = [ \({ 2 + \sqrt 2} \over2 \)]

Q4) \( 1 \over{ \sqrt{ 7}} \) = [ \(\sqrt{7}\over{ 7} \)]

Q4) \( 5 \over{ \sqrt{ 6}} \) = [ \(5\sqrt{6}\over{6} \)]

Q4) \({1} \over{ 4 + \sqrt 3} \) = [ \({ 4 - \sqrt 3} \over13 \)]

Q5) \( 1 \over{ \sqrt{ 10}} \) = [ \(\sqrt{10}\over{ 10} \)]

Q5) \( 2 \over{ \sqrt{ 5}} \) = [ \(2\sqrt{5}\over{5} \)]

Q5) \({1} \over{ 3 - \sqrt 3} \) = [ \({ 3 + \sqrt 3} \over6 \)]