Question

rationlise the denominator of √3+√2/5+√2​

Answer 1

Answer:

Rationalising Factor of[ (√3+√2)-√5]=[(√3+√2)+√5)]

Now Multiply Numerator and Denominator.

\begin{gathered}\frac{1( \sqrt{3 + } \sqrt{2} ) + \sqrt{5} }{[( \sqrt{3} + \sqrt{2} ) - \sqrt{5}]\:[ ( \sqrt{3} + \sqrt{2}) + \sqrt{5} ]} \\ \\ \frac{( \sqrt{3} + \sqrt{2} ) + \sqrt{5} }{( \sqrt{3} + \sqrt{2} ) {}^{2} - \sqrt{5} {}^{2} } \\ \\ \frac{ \sqrt{3} + \sqrt{2} + \sqrt{5} }{3 + 2 + 2 \sqrt{3 \times 2} - 5} \\ \\ \frac{ \sqrt{3} +\sqrt{2}+ \sqrt{5} }{5 + 2 \sqrt{6} - 5 } \\ \\ \frac{ \sqrt{3} + \sqrt{2} +\sqrt{5} }{2 \sqrt{6} }\end{gathered}

[(

3

+

2

)−

5

][(

3

+

2

)+

5

]

1(

3+

2

)+

5

(

3

+

2

)

2

−

5

2

(

3

+

2

)+

5

3+2+2

3×2

−5

3

+

2

+

5

5+2

6

−5

3

+

2

+

5

2

6

3

+

2

+

5

Since √6is an Irrational Number in Denominator For Make it Rational Multiply √6with both Numerator and Denominator.

\begin{gathered}\frac{ (\sqrt{3} + \sqrt{2}+ \sqrt{5} ) \times \sqrt{6} }{2 \sqrt{6} \times \sqrt{ 6 } } \\ \\ \frac{ \sqrt{6 \times 3} +\sqrt{6\times 2}+ \sqrt{5 \times6 } }{2 \times 6} \\ \\ \frac{ \sqrt{18} + \sqrt{12}+ \sqrt{30}}{12} \\ \\ \frac{ 3\sqrt{2} + 2\sqrt{3}+\sqrt{30}}{12}\end{gathered}

2

6

×

6

(

3

+

2

+

5

)×

6

2×6

6×3

+

6×2

+

5×6

12

18

+

12

+

30

12

3

2

+2

3

+

30

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Answer 2

Answer:

$= \frac{5 \sqrt{3} + 5 \sqrt{2} - \sqrt{6} - 2 }{23}$

Step-by-step explanation:

GIVEN,

$\frac{ \sqrt{3 } + \sqrt{2} }{5 + \sqrt{2} }$

RATIONALIZING DENOMINATOR;

$\frac{ \sqrt{3 } + \sqrt{2} }{5 + \sqrt{2} } \times \frac{5 - \sqrt{2} }{5 - \sqrt{2} }$

$= \frac{ (\sqrt{3 } + \sqrt{2})(5 - \sqrt{2}) }{(5 + \sqrt{2})(5 - \sqrt{2} ) }$

$= \frac{5 \sqrt{3 } - \sqrt{6} + 5 \sqrt{2} - \sqrt{4} }{ ({5})^{2} - {( \sqrt{2} })^{2} }$

$= \frac{5 \sqrt{3} + 5 \sqrt{2} - \sqrt{6} - 2 }{25 - 2}$

$= \frac{5 \sqrt{3} + 5 \sqrt{2} - \sqrt{6} - 2 }{23}$

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