Autorzy:Wojciech Babiański, Lech Chańko, Joanna Czarnowska
Wydawnictwo:Nowa Era
Rok wydania:2015
Usuń niewymierność z mianownika 4.71 gwiazdek na podstawie 7 opinii

Usuń niewymierność z mianownika

49Zadanie
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`a)\ 9/(5sqrt3)=9/(5sqrt3)*sqrt3/sqrt3=(strike9^3sqrt3)/(5*strike3^1)=(3sqrt3)/5` 

`b)\ 2/(sqrt3-1)=2/(sqrt3-1)*(sqrt3+1)/(sqrt3+1)=(2(sqrt3+1))/((sqrt3-1)(sqrt3+1))=(2(sqrt3+1))/(sqrt3^2-1^2)=(2(sqrt3+1))/(3-1)=(2(sqrt3+1))/(2)=sqrt3+1` 

`c)\ sqrt2/(3+sqrt2)=sqrt2/(3+sqrt2)*(3-sqrt2)/(3-sqrt2)=(sqrt2(3-sqrt2))/((3+sqrt2)(3-sqrt2))=(3sqrt2-2)/(3^2-sqrt2^2)=(3sqrt2-2)/(9-2)=(3sqrt2-2)/7` 

`d)\ (7sqrt2)/(2sqrt2-3)=(7sqrt2)/(2sqrt2-3)*(2sqrt2+3)/(2sqrt2+3)=(7sqrt2(2sqrt2+3))/((2sqrt2-3)(2sqrt2+3))=(7sqrt2*2sqrt2+7sqrt2*3)/((2sqrt2)^2-3^2)=(14*2+21sqrt2)/(4*2-9)=(28+21sqrt2)/(-1)=-28-2sqrt2` 

`e)\ (sqrt5+1)/(sqrt5-1)=(sqrt5+1)/(sqrt5-1)*(sqrt5+1)/(sqrt5+1)=(sqrt5+1)^2/((sqrt5-1)(sqrt5+1))=(sqrt5^2+2*sqrt5*1+1^2)/(sqrt5^2-1^2)=(5+2sqrt5+1)/(5-1)=(6+2sqrt5)/4=(3+sqrt5)/2` 

`f)\ sqrt6/(sqrt+sqrt3)=sqrt6/(sqrt2+sqrt3)*(sqrt2-sqrt3)/(sqrt2-sqrt3)=(sqrt6(sqrt2-sqrt3))/((sqrt2+sqrt3)(sqrt2-sqrt3))=(sqrt12-sqrt18)/(sqrt2^2-sqrt3^2)=(sqrt4*sqrt3-sqrt9*sqrt2)/(2-3)=(2sqrt3-3sqrt2)/(-1)=3sqrt2-2sqrt3` 

`g)\ (sqrt3-1)/(2sqrt3+sqrt2)=(sqrt3-1)/(2sqrt3+sqrt2)*(2sqrt3-sqrt2)/(2sqrt3-sqrt2)=((sqrt3-1)(2sqrt3-sqrt2))/((2sqrt3+sqrt2)(2sqrt3-sqrt2))=(sqrt3(2sqrt3-sqrt2)-1(2sqrt3-sqrt2))/((2sqrt3)^2-sqrt2^2)=` 

`\ \ \ =(2*3-sqrt6-2sqrt3+sqrt2)/(4*3-2)=(6-sqrt6-2sqrt3+sqrt2)/(12-2)=(6-sqrt6-2sqrt3+sqrt2)/10`