[3+root2][2-root3][3-root2][2+root3]
help me solve his...
Answers
Answer:
Heya there !!!
Here is the answer you were looking for:
\begin{lgathered}\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{lgathered}
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{lgathered}= \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{lgathered}
=
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
Answer:
hence answer is 7