Question

simplify 1 / √5+√6-√11​

Answer 1

Question :-

$\bf\dfrac{1}{ \sqrt{5} + \sqrt{6} - \sqrt{11} }$

We know, for Rationalising the Denominator of a Number, we firstly need to take out the Conjugate of the Denominator of the given Number.

Therefore, the conjugate of $\rm \sqrt{5} + \sqrt{6} - \sqrt{11}$ is $\rm (\sqrt{5} + \sqrt{6}) + (\sqrt{11})$

After, we get the Conjugate of the denominator we finally multiply it by the denominator and numerator.

$\sf{ \implies\dfrac{1}{( \sqrt{5} + \sqrt{6} ) - ( \sqrt{11}) } \times \dfrac{( \sqrt{5} + \sqrt{6} ) + ( \sqrt{11} )}{ ( \sqrt{5} + \sqrt{6} ) + ( \sqrt{11} ) } }$

$\bf{ \implies \dfrac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{( \sqrt{6} + \sqrt{5} ) ^{2} - ( \sqrt{11} ) ^{2} } }$

Using (a+b)² = a² + 2 × a × b + b²

$\bf \implies \dfrac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{6 + 5 + 2 \times \sqrt{6} \times \sqrt{5} - 11 }$

$\bf \implies \dfrac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{ \cancel{11 } - \cancel{11} + 2 \sqrt{30} }$

$\bf \implies \dfrac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{ 2 \sqrt{30} }$

We got the answer of the Rationalisation of $\bf\dfrac{1}{ \sqrt{5} + \sqrt{6} - \sqrt{11} }$

Once again, we will Ratitionalise it

Therefore, thus time we will Multiply the denominator and numerator by $\bf \sqrt{30}$

$\bf \implies \dfrac{ \sqrt{5} + \sqrt{6} + \sqrt{11} }{2 \sqrt{30} } \times \dfrac{ \sqrt{30} }{ \sqrt{30} }$

$\bf \implies \dfrac{ \sqrt{5 \times 30} + \sqrt{6 \times 30} + \sqrt{11 \times 30} }{2( \sqrt{30) ^{2} } }$

$\bf \implies \dfrac{\sqrt{5 \times 5 \times 6} + \sqrt{6 \times 6 \times 5} + \sqrt{ 330} }{2 \times 30}$

$\bf \implies \dfrac{5 \sqrt{6} + 6 \sqrt{5} + \sqrt{330} } {60}$

Answer 2

We have been asked to simplify $\tt{ \dfrac{1}{\sqrt{5} + \sqrt{6} - \sqrt{11}}}$ . To simplify it ; we need to rationalise the denominator. For rationalising the denominator and we need to take the opposite form of the the given fractions denominator and multiply it with numerator as well as the denominator.

Here the denominator is : $\tt{\sqrt{5} + \sqrt{6} - \sqrt{11}}$ So , to change its form; we will just change the sings. After we change the signs we get : $\tt{\sqrt{5} - \sqrt{6} + \sqrt{11}}$

Further we will multiply it with both numerator and denominator. So we get :

$\longrightarrow{\tt{\dfrac{1}{\sqrt{5} + \sqrt{6} - \sqrt {11}} \times \dfrac{\sqrt{5} - \sqrt{6} + \sqrt{11}}{\sqrt{5} - \sqrt{6} + \sqrt{11}}}}$

Multiplying the numerator as well as the denominator; we get :

$\longrightarrow{\tt{ \dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{{(\sqrt{5})}^{2} - {(\sqrt{6}}^{2} + {(\sqrt{11}}^{2}} }}$

The denominator can be written as :

$\longrightarrow{\tt{\dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{{(\sqrt{5} - \sqrt{6})}^{2} + {\sqrt{11}}^{2}}}}$

If we carefully observe the denominator is in the form of identity : $\tt{ {(a+b)}^{2} = {(a)}^{2} + {(b)}^{2} + 2ab}$

Following the same identity we will further solve this.

$\longrightarrow{\tt{\dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{{\sqrt{5}}^{2} + {\sqrt{6}}^{2} + {\sqrt{11}}^{2} + 2 \times \sqrt{6} \times \sqrt{5} }}}$

Thus we get :

$\longrightarrow{\tt{\dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{5 + 6 - 11 + 2\sqrt{30}}}}$

Further simplification gives :

$\longrightarrow{\tt{\dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{ 2\sqrt{30}}}}$

Now thoughwe have found the simpler denominator we still have to rationalize it.

Following the same method :

$\longrightarrow{\tt{\dfrac{\sqrt{5} + \sqrt{6} - \sqrt {11}}{2\sqrt{30}} \times \dfrac{\sqrt{30}}{\sqrt{30}}}}$

$\longrightarrow{\tt{\dfrac{(\sqrt{5 \times 30 } ) + \sqrt{6 \times 30} - \sqrt {11 \times 30}}{2{\sqrt{30}}^{2}}}}$

$\longrightarrow{\tt{\dfrac{\sqrt{5 \times 5 \times 6 } + \sqrt{6 \times 6 \times 5} - \sqrt{330}}{2\times30}}}$

$\longrightarrow{\tt{\dfrac{5\sqrt{6} + 6\sqrt{5} - \sqrt{330}}{60}}}$

Hence , simplified.