Question

Избавьтесь от иррациональности в знаменателе:1:1+корень из 2+ корень из 3

Answer 1

 

$frac{1}{1+sqrt{2}+sqrt{3}} = frac{1-sqrt{2}-sqrt{3}}{(1+sqrt{2}+sqrt{3})(1-sqrt{2}-sqrt{3})} = frac{1-sqrt{2}-sqrt{3}}{(1+(sqrt{2}+sqrt{3}))(1-(sqrt{2}+sqrt{3}))} =\\ frac{1-sqrt{2}-sqrt{3}}{1-(sqrt{2}+sqrt{3})^2} = frac{1-sqrt{2}-sqrt{3}}{1-(2+2sqrt{2}sqrt{3}+3)} = frac{1-sqrt{2}-sqrt{3}}{-4-2sqrt{2}sqrt{3}} =$

 

$frac{(1-sqrt{2}-sqrt{3})(4-2sqrt{2}sqrt{3})}{-(4+2sqrt{2}sqrt{3})(4-2sqrt{2}sqrt{3})} = frac{(1-sqrt{2}-sqrt{3})(4-2sqrt{2}sqrt{3})}{-(16-4*2*3)} =\\ frac{(1-sqrt{2}-sqrt{3})(4-2sqrt{2}sqrt{3})}{8} = frac{4 - 2sqrt{6} - 4sqrt{2} + 4sqrt{3} - 4sqrt{3} + 6sqrt{2}}{8} = frac{4 - 2sqrt{6} + 2sqrt{2}}{8}$

 

Answer 2

1/(1+√2+√3)=(4-2√6-4√2+4√3-4√3+6√2)/8=(4-2√6+2√2)/8