Question

rationalise the denominator of 1 / 1 + root2-root3

Answer 1

Answer:

After rationalising the given denominator, we get the answer as

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

Explanation:

The given expression is,

$\frac { 1 } { ( 1 + \sqrt { 2 } - \sqrt { 3 } ) }$

Rationalising the denominator

$\begin{array} { c } { \frac { 1 } { ( 1 + \sqrt { 2 } ) - \sqrt { 3 } } \times \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { ( 1 + \sqrt { 2 } ) + \sqrt { 3 } } } \\\\ { = \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { \left( ( 1 + \sqrt { 2 } ) ^ { 2 } - ( \sqrt { 3 } ) ^ { 2 } \right) } } \end{array}$

(As $(a^2-b^2) = (a+b)(a-b)$ )

$\begin{aligned} = & \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { ( ( 1 + 2 + 2 \sqrt { 2 } ) - 3 ) } \\\\ = & \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { 3 + 2 \sqrt { 2 } - 3 } \\\\ & = \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { 2 \sqrt { 2 } } \end{aligned}$

Multiply and divide by $\sqrt{2}$

$= \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { 2 \sqrt { 2 } } \times \frac { \sqrt { 2 } } { \sqrt { 2 } }$

After rationalising the given denominator, we get the answer as

$= \frac { \sqrt { 2 } + 2 + \sqrt { 6 } } { 4 }$

Answer 2

Answer:

this is answer u r question