Math, asked by randomcook123, 11 months ago

(2\sqrt{6} /\sqrt{2} +\sqrt{3} ) + (6\sqrt{2} /\sqrt{6} +\sqrt{3} )

Answers

Answered by Abdulrazak182
2

Answer:

by rationalizing the denominators we will get.

 \frac{2 \sqrt{6} }{ \sqrt{2}  +  \sqrt{3} }  \times  \frac{ \sqrt{2 } -  \sqrt{3}  }{ \sqrt{2}  -  \sqrt{3} }  \\  = \frac{2 \sqrt{12} - 2 \sqrt{18 }  }{2 - 3}  \\  =  \frac{4 \sqrt{3}  - 6 \sqrt{2} }{ - 1}  \\  = 6 \sqrt{2 }  - 4 \sqrt{3}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: (1)

by rationalizing second denominator, we get,

 \frac{6 \sqrt{2} }{ \sqrt{6} +  \sqrt{3}  } \times  \frac{ \sqrt{6}  -  \sqrt{3} }{ \sqrt{6} -  \sqrt{3}  }   \\   =  \frac{6 \sqrt{12}  - 6 \sqrt{6} }{6 - 3}  \\  =  \frac{6( \sqrt{12} -  \sqrt{6} ) }{3} \\  = 2(2 \sqrt{3} -  \sqrt{6}) \\  = 4 \sqrt{3}   - 2 \sqrt{6}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: (2) \\

adding (1) and (2),

6 \sqrt{2}  - 4 \sqrt{3}  + 4 \sqrt{3}  - 2 \sqrt{6}  \\  = 6 \sqrt{2}  - 2 \sqrt{6}  \\  = 2(3 \sqrt{2}  -  \sqrt{6} )

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