Question

plz solve Q-1,2,3,4.

Answer 1

Hope u like my process
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1)
Already answered in your next question.

2)
$\frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } + \frac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{ \sqrt{6}( \sqrt{6} + \sqrt{3}) + 3 \sqrt{2}( \sqrt{2} + \sqrt{3} ) }{( \sqrt{2} + \sqrt{3})( \sqrt{6} + \sqrt{3}) } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{6 + 3 \sqrt{2} + 6 + 3 \sqrt{6} }{2 \sqrt{3} + 3 \sqrt{2} + \sqrt{6} + 3 } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{(12 + 3 \sqrt{2} + 3 \sqrt{6})( \sqrt{6} + \sqrt{2}) - 4 \sqrt{3}(3 + 2 \sqrt{3} + 3 \sqrt{2} + \sqrt{6} ) }{(2 \sqrt{3} + 3 \sqrt{2} + \sqrt{6} + 3)( \sqrt{6} + \sqrt{2} ) } \\ \\ = \frac{12 \sqrt{6} + 6 \sqrt{3} + 18 + 12 \sqrt{2} + 6 + 6 \sqrt{3 } - 12 \sqrt{3} - 24 -12 \sqrt{6} - 12 \sqrt{2} }{6 \sqrt{2} + 6 \sqrt{3} + 6 + 3 \sqrt{6} + 2 \sqrt{6 } + 6 + 2 \sqrt{3} + 3 \sqrt{2} } \\ \\ = \frac{0}{9 \sqrt{2} + 8 \sqrt{3} + 5 \sqrt{6} + 12 } \\ \\ = 0$

3)
$3 \sqrt{45} - \sqrt{125} + \sqrt{200} - \sqrt{50} \\ \\ = 3 \sqrt{ {3}^{2} \times 5 } - \sqrt{ {5}^{2} \times 5 } + \sqrt{ {5}^{2} \times {2}^{2} \times 2} - \sqrt{ {5}^{2} \times 2 } \\ \\ = 3 \times 3 \sqrt{5} - 5 \sqrt{5} + 5 \times 2 \sqrt{2} - 5 \sqrt{2} \\ \\ = 9 \sqrt{5} - 5 \sqrt{5} + 10 \sqrt{2} - 5 \sqrt{5} \\ \\ = 4 \sqrt{5} + 5 \sqrt{2}$
4)
$\frac{4}{ {(216)}^{ \frac{ - 2}{3} } } - \frac{1}{ {(256)}^{ \frac{ - 3}{4} } } \\ \\ = 4 \times {(216)}^{ \frac{ 2}{3} } - {(256)}^{ \frac{3}{4} } \\ \\ = 4 \times { (\sqrt[3]{216}) }^{2} - {( \sqrt[4]{256}) }^{3} \\ \\ = 4 \times {(6)}^{2} - {(4)}^{3} \\ \\ = 144 - 64 \\ = 80$
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