Rationalise the denominator of each of the followingplz give the solut

1.	Rationalise the denominator of each of the followingplz give the solution fast
Answer» $\large\underline{\bf{Solution-}}$ Answer :- i $\red{\bf :\longmapsto\:\dfrac{4}{3 + \sqrt{3} } }$ Multiply and Divide by CONJUGATE of denominator, we GET $\: \: \rm = \: \: \dfrac{4}{3 + \sqrt{3}} \times \dfrac{3 - \sqrt{3} }{3 - \sqrt{3} }$ $\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{ {3}^{2} - {( \sqrt{3}) }^{2} }$ $\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$ $\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{9 - 3}$ $\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{6}$ $\: \: \rm = \: \: \dfrac{2(3 - \sqrt{3})}{3}$ $\: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\: \dfrac{4}{3 - \sqrt{3} } = \: \: \dfrac{2(3 - \sqrt{3})}{3}}}$ Answer :- ii $\green{\bf :\longmapsto\:\dfrac{5}{ \sqrt{2} + \sqrt{3} } }$ $\: \: \rm = \: \: \dfrac{5}{ \sqrt{3} + \sqrt{2} }$ Multiply and Divide by conjugate of denominator, we get $\: \: \rm = \: \: \dfrac{5}{ \sqrt{3} + \sqrt{2}} \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$ $\: \: \rm = \: \: \dfrac{5( \sqrt{3} - \sqrt{2}) }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} )}^{2} }$ $\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$ $\: \: \rm = \: \: \dfrac{5( \sqrt{3} - \sqrt{2})}{3 - 2}$ $\: \: \rm = \: \: 5( \sqrt{3} - \sqrt{2})$ $\: \: \: \: \: \: \: \green{ \boxed{\bf\implies \:\: \dfrac{5}{ \sqrt{3} + \sqrt{2} } = \: \: 5( \sqrt{3} - \sqrt{2})}}$ Answer :- iii $\blue{\bf :\longmapsto\:\dfrac{6 \sqrt{5} - 5 \sqrt{3} }{2 \sqrt{3} + 4 \sqrt{5}}}$ $\: \: \rm = \: \: \dfrac{6 \sqrt{5} - 5 \sqrt{3} }{4 \sqrt{5} + 2 \sqrt{3} }$ Multiply and Divide by conjugate of denominator, we get $\: \: \rm = \: \: \dfrac{6 \sqrt{5} - 5 \sqrt{3} }{4 \sqrt{5} + 2 \sqrt{3} } \times \dfrac{4 \sqrt{5} - 2 \sqrt{3} }{4 \sqrt{5} - 2 \sqrt{3} }$ $\: \: \rm = \: \: \dfrac{120 - 12 \sqrt{15} - 20 \sqrt{15} + 30}{ {(4 \sqrt{5} )}^{2} - {(2 \sqrt{3} )}^{2} }$ $\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$ $\: \: \rm = \: \: \dfrac{150 - 32 \sqrt{15} }{80 - 12}$ $\: \: \rm = \: \: \dfrac{150 - 32 \sqrt{15} }{68}$ $\: \: \rm = \: \: \dfrac{75 - 16 \sqrt{15} }{34}$ $\: \: \: \: \: \: \blue{ \boxed{\bf :\implies\:\dfrac{6 \sqrt{5} - 5 \sqrt{3} }{2 \sqrt{3} + 4 \sqrt{5}} = \dfrac{75 - 16 \sqrt{15} }{34} }}$ Additional Information More Identities to KNOW: (a + b)² = a² + 2ab + b² (a - b)² = a² - 2ab + b² a² - b² = (a + b)(a - b) (a + b)² = (a - b)² + 4ab (a - b)² = (a + b)² - 4ab (a + b)² + (a - b)² = 2(a² + b²) (a + b)³ = a³ + b³ + 3ab(a + b) (a - b)³ = a³ - b³ - 3ab(a - b)

Rationalise the denominator of each of the followingplz give the solution fast

Answer»

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

Answer :- i

$\red{\bf :\longmapsto\:\dfrac{4}{3 + \sqrt{3} } }$

Multiply and Divide by CONJUGATE of denominator, we GET

$\: \: \rm = \: \: \dfrac{4}{3 + \sqrt{3}} \times \dfrac{3 - \sqrt{3} }{3 - \sqrt{3} }$

$\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{ {3}^{2} - {( \sqrt{3}) }^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$

$\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{9 - 3}$

$\: \: \rm = \: \: \dfrac{4(3 - \sqrt{3})}{6}$

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

$\: \: \: \: \: \: \: \red{ \boxed{\bf\implies \:\: \dfrac{4}{3 - \sqrt{3} } = \: \: \dfrac{2(3 - \sqrt{3})}{3}}}$

Answer :- ii

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

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

Multiply and Divide by conjugate of denominator, we get

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

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

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$

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

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

$\: \: \: \: \: \: \: \green{ \boxed{\bf\implies \:\: \dfrac{5}{ \sqrt{3} + \sqrt{2} } = \: \: 5( \sqrt{3} - \sqrt{2})}}$

Answer :- iii

$\blue{\bf :\longmapsto\:\dfrac{6 \sqrt{5} - 5 \sqrt{3} }{2 \sqrt{3} + 4 \sqrt{5}}}$

$\: \: \rm = \: \: \dfrac{6 \sqrt{5} - 5 \sqrt{3} }{4 \sqrt{5} + 2 \sqrt{3} }$

Multiply and Divide by conjugate of denominator, we get

$\: \: \rm = \: \: \dfrac{6 \sqrt{5} - 5 \sqrt{3} }{4 \sqrt{5} + 2 \sqrt{3} } \times \dfrac{4 \sqrt{5} - 2 \sqrt{3} }{4 \sqrt{5} - 2 \sqrt{3} }$

$\: \: \rm = \: \: \dfrac{120 - 12 \sqrt{15} - 20 \sqrt{15} + 30}{ {(4 \sqrt{5} )}^{2} - {(2 \sqrt{3} )}^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$

$\: \: \rm = \: \: \dfrac{150 - 32 \sqrt{15} }{80 - 12}$

$\: \: \rm = \: \: \dfrac{150 - 32 \sqrt{15} }{68}$

$\: \: \rm = \: \: \dfrac{75 - 16 \sqrt{15} }{34}$

$\: \: \: \: \: \: \blue{ \boxed{\bf :\implies\:\dfrac{6 \sqrt{5} - 5 \sqrt{3} }{2 \sqrt{3} + 4 \sqrt{5}} = \dfrac{75 - 16 \sqrt{15} }{34} }}$

Additional Information

More Identities to KNOW:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)