Math, asked by Manjuprajapati2, 1 year ago

solved question 1 2 3 and 4

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Answered by rakeshmohata
0
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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