Question

Simplify
(Step by Step)

Answer 1

Answer :

Given :

$\longrightarrow{\frac{6}{2\sqrt {3} - \sqrt {6}} + \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} - {\frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}}}}$

Required to find :

1. Simplification of the given numbers.

Solution :

Let's Slove this question by splitting it into 3 parts .

1st part :

$\longrightarrow{ \frac{6}{2\sqrt {3} - \sqrt {6}}}$

Rationalising factor

$\boxed{2\sqrt {3} - \sqrt {6}}$

Hence; Now Rationalise the denominator.

$\longrightarrow{ \frac{6}{2\sqrt {3} - \sqrt {6}} \times \frac{2\sqrt {3}+\sqrt {6}}{2\sqrt {3} + \sqrt {6}}}$

$\longrightarrow{ \frac{6 (2\sqrt {3} - \sqrt {6})}{{(2\sqrt {3})}^{2} - {(\sqrt {6})}^{2}}}$

$\longrightarrow{ \frac{6 (2\sqrt {3} + 6 \sqrt {6})}{12 - 6}}$

$\longrightarrow{ \frac{6 (2\sqrt {3} + 6 \sqrt {6})}{6}}$

6 gets cancelled in numerator and denominator.

$\Rightarrow{ \frac{2\sqrt {3} + \sqrt {6}}{1}}$

Hence,

$\implies{2\sqrt {3} + \sqrt {6}}$

2nd part :

$\longrightarrow{ \frac{\sqrt {6}}{\sqrt {3} + \sqrt {2}}}$

Rationalising factor :

$\boxed{ \sqrt {3} - \sqrt {2}}$

Hence, Now Rationalise the denominator.

$\longrightarrow{ \frac{\sqrt {6}}{\sqrt {3} + \sqrt {2}} \times \frac{\sqrt {3} - \sqrt {2}}{\sqrt {3} - \sqrt {2}}}$

$\longrightarrow{ \frac{\sqrt{6}(\sqrt {3} - \sqrt {2})}{{(\sqrt {3})}^{2} - {(\sqrt {2}}^{2}}}$

$\longrightarrow{\frac{\sqrt {18} - \sqrt {12}}{3 - 2}}$

$\longrightarrow{\sqrt {9 \times 2} - \sqrt {4 \times 3}}$

Hence;

$\implies{3\sqrt {2} - 2\sqrt {3}}$

3rd part :

$\longrightarrow{\frac{4\sqrt {3}}{\sqrt {6} - \sqrt {2}}}$

Rationalising factor :

$\boxed{\sqrt {6} + \sqrt {2} }$

Hence Now; Rationalise the denominator .

$\longrightarrow{\frac{4\sqrt {3}}{\sqrt {6} - \sqrt {2}} \times \frac{\sqrt {6} + \sqrt {2}}{\sqrt {6} + \sqrt {2}}}$

$\longrightarrow{\frac{4\sqrt {3} ( \sqrt {6} + \sqrt {2})}{{(\sqrt {6})}^{2} - {(\sqrt {2})}^{2}}}$

$\longrightarrow{\frac{4 \sqrt {18} + 4 \sqrt {6}}{6 - 2}}$

$\longrightarrow{\frac{4 \sqrt {18} + 4 \sqrt {6}}{4}}$

$\longrightarrow{\frac{4 (\sqrt {18} + \sqrt {6})}{4}}$

$\longrightarrow{\frac{\cancel {4}(\sqrt{18 } + \sqrt {6})}{\cancel {4}}}$

( 4 gets cancelled in both numerator and denominator. )

$\Rightarrow{\sqrt {9 \times 2} + \sqrt {6}}$

Hence;

$\implies{3 \sqrt {2} + \sqrt {6}}$

According to the problem;

Now integrate the 3 parts.

$\small{\Rightarrow{2\sqrt {3}+\sqrt {6}+3\sqrt {2}-2\sqrt {3} - ( 3\sqrt {2} + \sqrt {6})}}$

$\small{\Rightarrow{2\sqrt {3}+\sqrt {6}+3\sqrt {2}-2\sqrt {3} - 3\sqrt {2} - \sqrt {6}}}$

$\small{\Rightarrow{3\sqrt {2} - 3\sqrt {2} +2\sqrt {3} - 2\sqrt {3} + 1\sqrt {6} - 1\sqrt {6}}}$

$\small{\Rightarrow{\cancel{3\sqrt {2} - 3\sqrt {2} +12\sqrt {3} - 2\sqrt {3} + 6\sqrt {6} - 1\sqrt {6}}}}$

$\boxed{ All \:numbers\: get\: cancelled \: due \:to\:opposite\:signs}$

$\implies{0}$

$\boxed{\therefore{value\:is\:0}}$

Answer 2

Question :

Simplify ,

$\frac{6}{2 \sqrt{3} - \sqrt{6} } + \frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{2} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} }$

Solution :

$\frac{6}{2 \sqrt{3} - \sqrt{6} } + \frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{2} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} }$

Now taking part 1 and rationalize the denominator,

$\frac{6}{2 \sqrt{3} - \sqrt{6} } \\ \\ = \frac{6(2 \sqrt{3} + \sqrt{6} ) }{(2 \sqrt{3} - \sqrt{6})(2 \sqrt{3} + \sqrt{6} ) } \\ \\ = \frac{6(2 \sqrt{3} + \sqrt{6} )}{ ({2 \sqrt{3}) }^{2} - { (\sqrt{6}) }^{2} } \\ \\ = \frac{6(2 \sqrt{3} + \sqrt{6} )}{12 - 6} \\ \\ = \frac{6(2 \sqrt{3} + \sqrt{6} )}{6} \\ \\ = 2 \sqrt{3} + \sqrt{6}$

Now taking 2nd part and rationalize the denominator,

$\frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ \sqrt{6}( \sqrt{3} - \sqrt{2} )}{ (\sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2)} } \\ \\ = \frac{ \sqrt{18} - \sqrt{12} }{ {( \sqrt{3}) }^{2} - { (\sqrt{2}) }^{2} } \\ \\ = \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 - 2} \\ \\ = 3 \sqrt{2} - 2 \sqrt{3}$

Now taking 3rd part and rationalize the denominator,

$\frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } \\ \\ = \frac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2} )}{( \sqrt{6} - \sqrt{2} )( \sqrt{6} + \sqrt{2}) } \\ \\ = \frac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2} )}{ {( \sqrt{6} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = \frac{4 \sqrt{3} ( \sqrt{6} + \sqrt{2} )}{6 - 2} \\ \\ = \frac{4 \sqrt{3}( \sqrt{6} + \sqrt{2} ) }{4} \\ \\ = \sqrt{18} + \sqrt{6} \\ \\ = 3 \sqrt{2} + \sqrt{6}$

Now integrate the 3 parts :

$(2 \sqrt{3} + \sqrt{6} ) + (3 \sqrt{2} - 2 \sqrt{3} ) - (3 \sqrt{2} + \sqrt{6} ) \\ \\ = 2 \sqrt{3} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3} - 3 \sqrt{2} - \sqrt{6} \\ \\ = 0$

Hence , value is 0.

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