rationalize the denominator of the expression 1/1+√6-√2
Answers
Answer:
Answer:
After rationalising the given denominator, we get the answer as
\frac { \sqrt { 2 } + 2 + \sqrt { 6 } } { 4 }
4
2
+2+
6
Explanation:
The given expression is,
\frac { 1 } { ( 1 + \sqrt { 2 } - \sqrt { 3 } ) }
(1+
2
−
3
)
1
Rationalising the denominator
\begin{gathered}\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}\end{gathered}
(1+
2
)−
3
1
×
(1+
2
)+
3
((1+
2
)+
3
)
=
((1+
2
)
2
−(
3
)
2
)
((1+
2
)+
3
)
(As (a^2-b^2) = (a+b)(a-b)(a
2
−b
2
)=(a+b)(a−b) )
\begin{gathered}\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}\end{gathered}
=
=
((1+2+2
2
)−3)
((1+
2
)+
3
)
3+2
2
−3
((1+
2
)+
3
)
=
2
2
((1+
2
)+
3
)
Multiply and divide by \sqrt{2}
2
= \frac { ( ( 1 + \sqrt { 2 } ) + \sqrt { 3 } ) } { 2 \sqrt { 2 } } \times \frac { \sqrt { 2 } } { \sqrt { 2 } }=
2
2
((1+
2
)+
3
)
×
2
2
After rationalising the given denominator, we get the answer as
= \frac { \sqrt { 2 } + 2 + \sqrt { 6 } } { 4 }=
4
2
+2+
6