Question

Simplify by rationalising the denominator 1/√11+√7-√8​

Answer 1

Step-by-step explanation:

\frac{1}{60}(6\sqrt{5}+5\sqrt{6}+\sqrt{330})

60

1

(6

5

+5

6

+

330

)

Step-by-step explanation:

Here, the given expression is,

\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}

6

+

5

−

11

1

For rationalizing the denominator, multiply both numerator and denominator by √6 + √5 + √11,

=\frac{1}{\sqrt{6}+\sqrt{5}-\sqrt{11}}\times \frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{\sqrt{6}+\sqrt{5}+\sqrt{11}}=

6

+

5

−

11

1

×

6

+

5

+

11

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{(\sqrt{6}+\sqrt{5})^2-(\sqrt{11})^2}=

(

6

+

5

)

2

−(

11

)

2

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{6+5+2\times\sqrt{6}\times \sqrt{5}-11}=

6+5+2×

6

×

5

−11

6

+

5

+

11

=\frac{\sqrt{6}+\sqrt{5}+\sqrt{11}}{2\sqrt{30}}=

2

30

6

+

5

+

11

Again for rationalizing the denominator, multiply both numerator and denominator by √30,

=\frac{\sqrt{180}+\sqrt{150}+\sqrt{330}}{60}=

60

180

+

150

+

330

=\frac{6\sqrt{5}+5\sqrt{6}+\sqrt{330}}{60}=

60

6

5

+5

6

+

330

=\frac{1}{60}(6\sqrt{5}+5\sqrt{6}+\sqrt{330})=

60

1

(6

5

+5

6

+

330

Answer 2

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

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{11} + \sqrt{7} - \sqrt{8}}$

$\rm \: \: = \: \dfrac{1}{ (\sqrt{11} + \sqrt{7}) - \sqrt{8}}$

☆ On rationalizing the denominator, we get

$\rm \: \: = \: \dfrac{1}{ (\sqrt{11} + \sqrt{7}) - \sqrt{8}} \times \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{( \sqrt{11} + \sqrt{7}) + \sqrt{8}}$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{ {( \sqrt{11} + \sqrt{7})}^{2} - {( \sqrt{8})}^{2} }$

$\: \: \: \: \: \: \: \: \green{\bigg \{ \bf \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{(11 + 7 + 2 \sqrt{77})- 8 }$

$\: \: \: \: \: \: \: \: \: \pink{\bigg \{ \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy\bigg \}}$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{18 + 2 \sqrt{77}- 8 }$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{10 + 2 \sqrt{77}}$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{2(5 + \sqrt{77})}$

$\rm \: \: = \: \dfrac{ \sqrt{11} + \sqrt{7} + \sqrt{8} }{2(5 + \sqrt{77})} \times \dfrac{5 - \sqrt{77} }{5 - \sqrt{77} }$

$\rm \: \: = \: \dfrac{5 \sqrt{11} +5 \sqrt{7} + 5 \sqrt{8} - 11 \sqrt{7} - 7 \sqrt{11} - 2 \sqrt{154} }{2\bigg( {5}^{2} - {( \sqrt{77} )}^{2} \bigg) }$

$\: \: \: \: \: \: \: \: \purple{\bigg \{ \bf \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: \: = \: \dfrac{10 \sqrt{2} - 2 \sqrt{154} - 6 \sqrt{7} - 2 \sqrt{11})}{2(25 - 77)}$

$\rm \: \: = \: \dfrac{2(5\sqrt{2} - \sqrt{154} - 3 \sqrt{7} - \sqrt{11}) }{2(25 - 77)}$

$\rm \: \: = \: \dfrac{(5\sqrt{2} - \sqrt{154} - 3 \sqrt{7} - \sqrt{11}) }{ - 52}$

$\rm \: \: = \: \dfrac{( - 5\sqrt{2} + \sqrt{154} + 3 \sqrt{7} + \sqrt{11}) }{ 52}$

$\rm \: \: = \: \dfrac{ \sqrt{154} + \sqrt{11} + 3 \sqrt{7} - 5 \sqrt{2} }{52}$

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)