Question

rationalize the denominator of √2÷√6-√2 ​

Answer 1

Step-by-step explanation:

this would be a correct answer bro

Answer 2

Answer:

${\sf{ {\dfrac{ {\sqrt{3}} + 1}{2}} }}$

Step-by-step explanation:

Given : ${\sf{\ \ {\dfrac{ {\sqrt{2}} }{ {\sqrt{6}} - {\sqrt{2}} }} }}$

$\Rightarrow{\sf{ {\dfrac{ {\sqrt{2}} }{ {\sqrt{6}} - {\sqrt{2}} }} \times {\dfrac{ {\sqrt{6}} + {\sqrt{2}} }{ {\sqrt{6}} + {\sqrt{2}} }} }}$

$\Rightarrow{\sf{ {\dfrac{ ( {\sqrt{2}} )( {\sqrt{6}} + {\sqrt{2}} )}{ ( {\sqrt{6}} - {\sqrt{2}} )( {\sqrt{6}} + {\sqrt{2}} ) }}}}$

${\boxed{\sf{\red{Identity \ : \ (a - b)(a + b) = a^2 - b^2}}}}$

${\sf{\red{Here, \ a = {\sqrt{6}} , \ b = {\sqrt{2}} }}}$

$\Rightarrow{\sf{ {\dfrac{ ( {\sqrt{2}} )( {\sqrt{6}} + {\sqrt{2}} )}{ ( {\sqrt{6}} )^2 - ( {\sqrt{2}} )^2 }}}}$

$\Rightarrow{\sf{ {\dfrac{ ( {\sqrt{2}} )( {\sqrt{6}} ) + ( {\sqrt{2}} )( {\sqrt{2}} ) }{ 6 - 2 }} }}$

$\Rightarrow{\sf{ {\dfrac{ {\sqrt{2 \times 2 \times 3}} + {\sqrt{2 \times 2}} }{ 4 }} }}$

$\Rightarrow{\sf{ {\dfrac{2 {\sqrt{3}} + 2}{4}} }}$

$\Rightarrow{\sf{ {\dfrac{2 ( {\sqrt{3}} + 1 )}{4}} }}$

$\Rightarrow{\boxed{\sf{\green{ {\dfrac{ {\sqrt{3}} + 1}{2}} }}}}$