Math, asked by enjooremmanuel, 8 hours ago

2 root 3 + 3 root 2 / 3 root 2 - 2 root 3

Answers

Answered by btbrainly
0

Answer:

1000

Step-by-step explanation:

colud be it but not sure

Answered by princydavid3879
2

Answer:

Here is the answer you were looking for:

\begin{gathered} \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } + \frac{ \sqrt{12} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } - \frac{ \sqrt{2 \times 2 \times 3} }{ \sqrt{3} - \sqrt{2} } \\ \\ = \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } - \frac{2 \sqrt{3} }{ \sqrt{3} - \sqrt{2} } \\ \end{gathered}

3

2

+2

3

3

2

−2

3

+

3

2

12

=

3

2

+2

3

3

2

−2

3

3

2

2×2×3

=

3

2

+2

3

3

2

−2

3

3

2

2

3

On rationalizing the denominators we get,

\begin{gathered} = \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} + 2 \sqrt{3} } \times \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 \sqrt{2} - 2 \sqrt{3} } - \frac{2 \sqrt{3} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = \frac{ {(3 \sqrt{2} )}^{2} + {(2 \sqrt{3} )}^{2} + 2(3 \sqrt{2} )(2 \sqrt{3}) }{ {(3 \sqrt{2}) }^{2} - {(2 \sqrt{3}) }^{2} } - \frac{2 \sqrt{3}( \sqrt{3} + \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = \frac{18 + 12 + 12 \sqrt{6} }{18 - 12} - \frac{6 - 2 \sqrt{6} }{3 - 2} \\ \\ = \frac{30 + 12 \sqrt{6} }{6} - 6 + 2 \sqrt{6} \\ \\ = 5 +2 \sqrt{6} - 6 + 2 \sqrt{6} \\ \\ = - 1 + 4 \sqrt{6} \\ \\ = 4 \sqrt{6} - 1\end{gathered}

=

3

2

+2

3

3

2

−2

3

×

3

2

−2

3

3

2

−2

3

3

2

2

3

×

3

+

2

3

+

2

=

(3

2

)

2

−(2

3

)

2

(3

2

)

2

+(2

3

)

2

+2(3

2

)(2

3

)

(

3

)

2

−(

2

)

2

2

3

(

3

+

2

)

=

18−12

18+12+12

6

3−2

6−2

6

=

6

30+12

6

−6+2

6

=5+2

6

−6+2

6

=−1+4

6

=4

6

−1

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