Question

Помогите освободить от иррациональности

$frac{1}{ sqrt{2} + sqrt{5}- sqrt{3} }$

$frac{6}{ sqrt{10}+ sqrt{6}+5- sqrt{15} }$

Answer 1

$1); ; frac{1}{sqrt2+sqrt5-sqrt3}= frac{sqrt2+sqrt5+sqrt3}{(sqrt2+sqrt5)^2-3}= frac{sqrt2+sqrt5+sqrt3}{7+2sqrt{10}-3} =frac{sqrt2+sqrt5+sqrt3}{2(sqrt{10}+2)}=\\= frac{(sqrt2+sqrt5+sqrt3)(sqrt{10}-2)}{2cdot (10-4)}= frac{(sqrt2+sqrt5+sqrt3)(sqrt{10}-2)}{12}=\\= frac{(sqrt2+sqrt5+sqrt3)(sqrt5-sqrt2)cdot sqrt2}{12}= frac{sqrt2cdot ((5-2)+sqrt3(sqrt5-sqrt2))}{12}=\\= frac{sqrt2(3+sqrt{15}-sqrt6)}{12}$

$2); ; frac{6}{sqrt{10}+sqrt6+5-sqrt{15}}=frac{6cdot ((sqrt{10}+sqrt6)-(5-sqrt{15}))}{(sqrt{10}+sqrt6)^2-(5-sqrt{15})^2}=frac{6cdot (sqrt{10}+sqrt6-5+sqrt{15})}{16+2sqrt{60}-40+10sqrt{15}}=\\= frac{6(sqrt{10}+sqrt6-5+ sqrt{15})}{14sqrt{15}-24}=frac{6(sqrt{10}+sqrt6-5+sqrt{15})}{2(7sqrt{15}-12)}=\\=frac{3(sqrt{10}+sqrt6-5+sqrt{15})(7sqrt{15}+12)}{735-144}=- frac{3(sqrt{10}+sqrt6-5+sqrt{15})(7sqrt5+12)}{591}$