Question

Освободиться от иррациональности в знаменателе дроба: 3+√2+√3 / 3-√2-√3

Answer 1

$= frac{(3+ sqrt{2}+ sqrt{3} )(3- sqrt{2} + sqrt{3} )}{(3- sqrt{2} - sqrt{3} )(3- sqrt{2} + sqrt{3} )} = frac{(3+ sqrt{3}+ sqrt{2})(3+ sqrt{3}- sqrt{2} ) }{(3- sqrt{2} )^2 -( sqrt{3} )^2}= \ \ = frac{(3+ sqrt{3} )^2-( sqrt{2} )^2}{9-6 sqrt{2}+2-3 }= frac{9+6 sqrt{3}+3-2 }{8-6 sqrt{2} }= frac{10+6 sqrt{3} }{8-6 sqrt{2} }= \ \ = frac{2(5+3 sqrt{3} )}{2(4-3 sqrt{2} )}= frac{5+3 sqrt{3} }{4-3 sqrt{2} }= \ \ = frac{(5+3 sqrt{3} )(4+3 sqrt{2} )}{(4-[tex] sqrt{2}(4+3 sqrt{2} )}= frac{20+12 sqrt{3}+15 sqrt{2}+9 sqrt{6} }{4^2-(3 sqrt{2} )^2}= \ \ = frac{20+12 sqrt{3}+15 sqrt{2}+9 sqrt{6} }{16-18}= -10 -6 sqrt{3} -7.5 sqrt{2}-4.5 sqrt{6}$)()} [/tex]